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UNIT 1 HIGHER ORDER DERIVATIVES
Structure
1.1 Introduction Objectives
1.2 Second and Third Order Derivatives 1.3 nth Order Derivatives 1.4 Leibniz Theorem 1.5 Taylors Series and Maclaurins Series 1.6 Summary 1.7 Solutions and Answers
1.1 INTRODUCTION
In the first block you have found the derivatives of a number of functions. You know that the derivative f of a differentiable function f is again a function and is called the derived function off.
We have already seen in Unit 3 that the concept of differentiation was motivated by some physical concepts (I~ke the velocity of a moving particle) and also by geometrical notions (like the slope of a tangent to a curve). The second and higher order derivatives are also sunilarly motivated by some physical considerations (like the acceleration) and some geometrical ideas (like the curvature of a curve), which we shall study in the remaining units of this block.
We shall introduce higher order derivatives in Sec. 1 and 2. Leibniz Theorem which is given in Sec. 4 gives us a formula for finding the higher derivatives of a product of two fufil~cjns In the later sections, we shall study some useful formulas, called series expansions. The sigir~ficance of these expansions will become clearer in Unit 1 of Block 4.
Objectives
After reading this unit you should be able to :
calculate higher order derivatives of a given h c t i o n f use the Leibniz formula to find the nth derivatives of products of functions expand a function using Taylor's Maclaurin's series.
1.2 SECOND AND THIRD ORDER DERIVATIVES ----- I
Consider the fhnction f(x) = x4. You know that f (x) = 4x3. Now, this f is again a polynomial
I fhnction and hence, can be differentiated (see Example 5, Unit 3). We shall denote the derivative o f f by f '. Thus.
f' (x) = 1 2x2. This f"' (x) is calledthe second derivative of the fhnction fat the point x. It is also
d 2~ denoted by (read as d square y by d x square ) of yz or e2) or ~ ' y .
dx
I - .
Let us differentiate f'. We get f" (x) = 24x, where Y'denotes the derivative off ' , or the third
d3y derivative off. Other notations of f" (x) are 7 or y, or e3), or D3y. Differeiitiating f"', dx
d4y we get the fourth derivative off, P4) (x) = 7 = y4 = 24. . dx
Thus, differentiating (if possible) a given function f, then differentiating its derivative, and then derivative's derivative and so on, we get the first derivative, the second derivative, the third, fourth, ........ derivatives of the function f.
dny If n is any positive integer, then the nh derivative o f f is denoted by f(") or by - (read as d.
dxn nybydxn)orbyy,orDny.
Drawing Curves
Such equaiions involving the derivatives are known as differential equations .
Recall (Unit I ) that
Note that in the notation 6") the bracket is necessary to distinguish it from f", that is, f raised to the power n. This process of differentiating again and again, in succession, is called sllccessive differentiation.
We have already seen that there are functions f that are not differentiable. In other words f need not always exist. Similarly even when f exists it is possible that f' does not exist (see ~xample 3 near the end of this section). In general, for each positive integer n there are functions f such that 6") exists, but itn+') does not exist. However, many functions that we consider in these sections possess all higher derivatives.
A twice differentiable function is a function f such that f' exists. ~ e t n be a positive integer. A function f such that ff") exists is called an n-times differentiable function. If $3 exists for every positive integer n, then f is said to be an infinitely differentiable function.
Now we give some simple examples of higher derivatives.
Example 1 If we are given that the third derivative of the function
f(x) = ax3 + bx + c has the value 6 at the point x = 1, can we find the value of a?
Here, f(x) = ax3 + bx + c
Differentiation this we get
f (x) = 3axZ + b
Differentiating this again, we get
P(x)=6ax
Differentiating once again, we get
P) (x) = 6a
Taking the value at x = 1,
P3)(1) - 6a
It is given that P ) ( l ) = 6.
Thus 6a = 6. Therefore a = 1.
Example 2 If y = 2 sin x + 3 cos x + 5, let us prove that yz + y = 5. Now,y=2sinx+3cosx+5
y, =2cosx-3sinx yz=-2sinx-3cosx :. yz+y=-2sinx-3cosx+ 2s inx+3cosx+5=5
The example below gives a function f for which f exists but f' does not exist.
Example 3 Consider,the function f(x) = x 1 x 1 for all x in R The furaction f(x) can be rewritten as
At points other than 0 we have f (x)=2xifx>O f (x)=-2xifx<O At x = 0, the right derivative off,
h2 - o2 - lim h = 0, and Rf (0) = lim+ - - h-bo h h+o+
the left derivative off,
h2 - o2 - lim h=O. Lf (0) = lim - - h-bo- h h-bo-
Therefore f (0) = 0. Thus,f (x)=2 IxIforallxinR
We have already seen in Example 7, Unit 3 that the absolute value furaction I x 1 fails to be differentiable at 0. Therefore, f is not differentiable at x = 0. That is, fc2) (0) does not exist.
some exercises before going any further.
Find the second derivatives of the following a) f(x)=x3-4 b) y = e2X
functions.
Higher Order Derivatives
, I
;.':
i.;:
& E2) Find f13) (7~4) for the following functions, a) fTx) =sec x b) f(x)=sin2x+cos2x
> t, 1. 5. ,
C .
E E3) 'Prove that the following functions satisfy the differential equations shown against them. a) y =sinx; Y4 = Y b) y=cosx ; (y2)* + ( ~ 3 ) ~ - 1
a
Drawing Curves E E4) Find the value of integer k in each of the following a) f(x) = sink x where fi2)(d6) = 2 J? b) f(x)=xk+kx2+ 1 wherefi2)(1)=12
1.3 nth ORDER DERIVATIVES . Let n be a natural number. We have already defined the n'h derivative of a hnction in Sec. 2.
When a function f is given by a formula, it is often necessary to express its nh derivative also by a formula using f and n. Usually, one can guess P") after working out $I), fi2) and P). However, a rigorous proof would require an application of the principle of mathematical induction.
In the example below we shall derive formulas for the nfi derivative of various functions. Study them carefully as we shall be using them in later sections.
But, first let us recall the principle of mathematical induction.
If {P,) is a sequence of propositions (statements) satisfying. 9 P, is true (usually N = 1). ii) The truth of Pi implies the truth of Pi+,, i 2 N,
then Pn is true for all n 2 N. 1 , We shall apply this principle in the examples that follow. 1 Example 4 We shall prove here that the n"' derivative of the polynomial hnction xn', m 2 0, w.r.t. x is
dnxm m(m-1) ....( m-n+l)xm-",if nSm,and
dx" 0, if n > m Let us denote by Pn the statement
Note that the product m (m- 1) ,..... (m- n + 1) has n factok. When n = 1, we have already shown that
Thus-thcstatement P, namely. (5 mm*m-l) is true.
Next, suppose we have proved for some n that the statement Pn is true. This means that the nh derivative of xm is
m(m - 1) .... m(m - n + I), xm-". if n I m, and is
0, if n > m
Then the (n + 1)" derivative of xm is f = the derivative of the nh derivative of xm
m (m - 1) ..... (m - n + I) (m - n: xm-"-I if n < m
Oif n > m.
! m(m-1) .....( m - n + I ) ( m - n ) x m-(n-1) ,if n + l < m
O i f n + I > m .
This means that the truth of P, implies the truth of P,,,. Therefore, by the principle of mathematical induction, P, is true for all n > I . Hence our result is true for all natural numbers n.
Remark 1 When n = m, the nth derivative of x'" is =m(m- 1) ....... (m-n+ 1)xn'"=m(m- 1) ......... 3.2.1.Thisisthesarne1sm!.
Example 5 If f(x) = In (1 + x), let us find e") (x). 1
Differentiating f ( x ) = In (1 + x), we get f (x) = - 1 + x
I Differentiating again, fc2)(x) =- -
(1 + x ) ~ 2
Differentiating once again, E3) (x) = - (1 +
Can you guess fin) (x) now? If you have guessed correctly, you must have arrived at these conclusions.
i) The denominator of fin) (x) is ( 1 + x)".
ii) Its sign is positive or negative according as n is odd or even.
iii) Its numerator has (n - I)! Do not think that it is merely (n - 1). There is a factorial symbol too. To be convinced of this, calculate E4) (x) and see.
(-I),-' x (n - I)! %
'Therefore our guess is fir#) (x) = (1 + x)" ...( I)
This guess remains to be proved. A proof is necessary because there could exist many other formulas for P") (x) that coincide with the correct answer when n = 1.2 or 3. For example, if we omit the factorial symbol, we get one such formula. But we have already mentioned that this formula does not hold for E4) (x). So, let's try to prove (I).
We first note that it is true for n = 1,2 and 3 as we have seen in the beginning.
Assume that it is true for n = m, that is,
(-I)"'-' x (m - I)! fi'") (x) =
(1 + x)"'
Differentiating this we get,
fil'l+l) (x) = (-1 y*-1 (m - I)! -m (-I)"'-' (-1) x m (m - I)! - (1 + X ) ~ + I - (1 + x)"'+~
Thls proves the guess for n = m t 1. Thus, assuming the truth of the formula (1) for n = m, we arrive at the huth of this formula for n = m + 1 . Therefore by the principle of mathematical inductioc, our guessed answer w correct for all positive integers n.
Thus, when f(x) = In (I + x).
(-I)"-' (11 - I ) ! fin'(x) =
(I + X)n
Higher Order Uerivat ibc*
When 11 .- m. the nlh C ~ ~ I I V . I ! I \ ,
IS cor~stdnt, because u" '. - x I , therefore the (n r 1 ) "
d e ~ ~ v a t ~ v e 1s /em
Check !hill IIIC i o l ~ t l ~ ~ l < ; n s - ' I ] -
nl or 13 > rn " and "n r i 2 n;"
are equlvalenL.
Drawing Curves
Recall that cos (9 + 112) = - sin 8 cos (0 + I) = - co* 8- cos ( + 3 ~ ~ 2 ) = -SIP 9
10
E E 5) If y = (1 + x)'; where r is a real number, find y,, where n is natural number (n < r).
Example 6 If f(x) = cos 2x, let us use the principle of mathematical induction to find a formula for 6") (0).
We first find fln)(x) when n = 1,2,3,4.
We have f(x) = cos 2x.
Differentiating this successively, we get fi') (x) = - 2 sin 2x fi2' (x) = - 4 cos 2x fi3) (x) = 8 sin 2x fi4)(x)=16c0s2x
We see that in the formula for fin) (x), we have to have
i) a sign (positive or negative),
ii) a coefficient (some power of 2), and
iii) a trigonometric function (sin 2x or cos 2x)
We observe that the first two terms cany negative sign, the next two carry positive sign, the next two negative and so on.
We also observe that sin and cos occur alternately. Therefore our guess is
I -2" sin 2x if n is the form 4k + 1 -2"cos 2x if n is of the form 4k + 2
f(") (x) = 2"ssin 2x if n is of the form 4k + 3 . . .(2)
(2"cos 2x jf n is of the form 4k
Wecanalso write this in a compact form as #@(x) = 211 cos (2x + nn / 2) ...( 3)
You can easily check that (2) and (3) are equivalent by putting n = 4k + 1,4k + 2,4k + 3 and 4k in (3). We shall now prove formula (3).
..
We have already seen that it is true for n = 1,2,3 and 4. Suppose it is true for n = m, that is, P) (x) = 2"' cos (2x + d 2 ) . Differentiating this we get, P+') (x) = 2"'.2.(- sin 2x + m d 2 ) sin (2x + m 7~12) or, Pnl+') (x) = 2"'+' cos [2x + (m + 1) d 2 ]
So here again we see that the truth of the formula (3) for n = m implies its truth for
I Therefore by the principle of mathematical induction, we see that the guessed formula for fin) (x) is true for all natural numbers n.
Now substitute x = 0. We obtain
I f(") (0) = 2" cos nn12
/ Th~r s the required answer.
We can also use this method to prove a general result for the nth derivative of a sum of two functions.
Example 7 Iff and g are two functions from R to R and ifboth of them are n-times differentiable, we can prove that
We shall prove this result by induction. When n = 1, this means (f + g)' = f + g'. This has already been proved in Unit 3. Suppose (f + g)(ln) = fin') + g(") is true.
Differentiating this we get [(f + g)(m)]' = [f("'J + g(m)]r = [f("')lp + [g(m)]r -
- I This is the same as (f+g)(m+l)=fI*I)+g(nl+l)
i Thus the result is true for n = m + 1. Therefore by the principle of mathematical Induction (f + E)(n) = P"+ g(ll) holds for all natural numbers n.
Remark 3 Similarly one can prove that (cf)(") = c.Pn) holds for all natural numbers n and all scalars c. This fact combined with Example 7 can be restated in the6'linear algebra terminology" as :
I ' The collection of n-times differentiate functions is a vector space under usual operations.
Try to solve these exercise now.
E E6) Find the n"' derivative of the following functions
a) f(x) = (ax + b)3
b) f(x) =(ax + b)"'
b c) f ( x ) = ~
Higher Order Derivatives
cos (2x + (m + 1 ) x12) = cos (2x + n1d2 +d2) = - sin (2x + rnd2)
Drawing Curves
E E8) If y = sin (ax + b), prove that for every positive integer n. we have y, = an sin [(nn/2) + ax t b]
E E 7 ) If f(x) = sin x, prove that fin) (x) = sin [x + nn/2] holds for every natural number n.
E E9) Prove that the n'" derivative of the polynomial function f (x) = a, + alx + a,x2 + ..... + anxl' is the constant function n! a,, .
E E 10) If y = cos x and if n is any positive integer, prove that [y,I2 + [yn+,I2 = 1-
1.4 LEIBNIZ THEOREM
In Unit 3 we have proved some rules regarding the derivatives of the sum, scalar multip!e, product and quotient of two differentiable functions. These were
(f+g) '=f +g' (cf)' = c f (fg) = fg' + g f
gf' - fg' (flg)' = ,2 (g (x) # 0 anywhere in the domain)
b
In the last section we have seen. (Example 7 and Remark 3) that the first two rules can be extended to the n"' derivatives i f f and g are n-times d~fferentiable functions. In this section we are going to extend the product rule of differentiation. We shall give a formula for the
! nLh derivative of the product of two functions. i
The product rule for two functions u and v can also be written as (uv)] = U I V + U V I
Now we look for a similar formula for (uv)?, (uv),, etc.
But first let us recall the meaning of the notation C(n, r), where n and r E Z+ and r 5 n. C(n, r) stand for the number of ways of choosing r objects from n objects. Socletinles it is also
denoted by I1Cr or ( : ) .
Drawing Curves Also recall the formulas
ii) C(n, 0) = C(n, n) = 1
iii) C(n, r) = C(n, n - r)
iv) C(n,r)+C(n,r+ l )=C(n+ l , r + 1)
These are combinatorial identities, true for all positive integers r and n with r 5 n.
Theorem 1 (Leibniz Theorem) Let n be a positive integer. If u and v are n times differentiable functions, then
(uv), = C(n, 0) u,v + C(n, 1) u,,v, + C(n, 2) uW2v2 + ....... + C (n, n) u v,
Leibniz had stated this result in The pattern in the formula for (uv), can be compared with the expansion of (x + y)". The his first article on differential coefficients are binomial coefficients and they appear in the same order as those in the
which was published in expansion of (x + y)". The order of the derivative of u goes on decreasing one at a time, and 1684. the order of the derivative of v goes on increasing one at a time. The number of terms is n + 1.
Remark 4 We omit the proof of this theorem and merely indicate how this can be proved by induction on n. Firstly, when n = 1, the above formula is the same as the already known product formula, and therefore is true. Assuming that it is true for n = m, we can prove it for n = m + 1, by applying the product rule for each term of the expansion of (uv), and by using the combinatorial identities mentioned. (See E 17) for more details.
We start with a simple and direct application of the formula.
Example 8 If f(x) = x sin x, let us find the fourth derivative off, using Leibniz Theorem.
We first observe that far n = 4, the Leibniz Theorem states
In this problem we take u = x and v = sin x, so that f = uv
We have u = x v=sinx u, = 1 v1 = COS X
%=o=ujr .u4 v2=-sinx v3=-COSX v, = sin x
Substituting these in the above formula, we get
What happens if we attack the same problem directly without the use of Leibniz Theorem? We have
f(x) =x sin x
Differentiating this, we get
f (x) = x cos x + sin x (by product rule)
Differentiating once again, we get
r ( x ) =x(-sinx)+ 1cosx+cosx =2cosx-xsinx
Differentiating once again, we get
fi3) (x) = -2 sin x - (x cos x + sin x) =-3sinx-xcosx
Differentiating once again,
fi4)(x) =-3cosx-[x(-sinx)+cosx] =xsinx-4cosx
We notice that we obtain the same answer. In this direct method, we had to apply the product formula four times, once for each differentiation.
It is clear that when we want the n" derivative for bigger values of n, Leibniz theorem provides an easier method to write down the answer, avoiding the difficulty of repeatedly applying the product formula.
Example 9 If y =(~in- 'x)~, prove that
(1 - X2) Y , + ~ - (2n + 1) xy,+, - n2y, = 0 for each positive integer n.
Differentiating both sides of y = sin-'^)^, we get
2 sin-' x
y1= ,/= Squaring and crossmultiplying we get ( 1 - ~ ~ ) ~ ~ ~ = 4 ( ~ 1 I 1 - I ~ ) ~ = 4 y
Differentiating once again, we get 2(1 -x2) yly2-2xyI2-4y, = o
Dividing throughout by 2 y, gives us (1-x2)y2-xyl-2=0
Differentiating n times, using Leibniz Theorem for each of the first two terms we get (1 -x2) y,,, -C(n, 1) 2xy,,, - C(n, 2) 2 Y,,- {xY,+, +C(n, 1) yn) =O
That is, ( I - x2) Y,,+~ - (2n + 1) xy,,, - n2y, = 0
The following exercises will give you some practice in applying Leibniz Theorem.
E 11) State Leibniz Theorem when n = 5. That is, (U.V)~ = ?
E 12) Prove that when n = 1, Leibniz Theorem reduces to the product rule of differentiation.
E 13) Find the third derivative of sin2x using Leibniz Theorem. Find the same directly also and verify that you obtain the same answer.
Hlgher Order Derivatives
Drawing Curves
E E 14) If f(x) = x eX, find the sixth derivative of f, using Leibniz fonnula.
' E E 15) Find thc nn derivative of x3 lnx
E E16) If y = F x2 prove that y(")=eU[a"x2+2na*' x+n(n- 1)aW2]
I E E 17) a) Write down Leibniz formula for (u.v), H i g h e r O r d e r Der iva t i ves
b) Differentiate it term by term and obtain (~v),~=C(m,O)u,~ v+C(m, 1)(u,,,vI+u,,v2)+ ........ +C(m,rn)uv,,.
c) Deduce that (UV),, =c(m,O)u,, v+ [C(111,0)+C(m, 1)1u,"v1 =LC(- I)+ C(m,2)]~,~ v2+ ......... +[C(m,m- 1)+C(m,rn)]u1v,+C(m,m)uv,,.
1 d) Deduce from (c) the Leibniz foxmula for (uv), I. I
I . i t i
E E 18) Using Leibniz Theorem and induction, prove that
1 (xn)(") = n! for all natural numbers n.
1.5 TAYLOR'S SERIES AND MRCLAURIN'S SERIES
In this section we obtain aeries expansions for many important functions. For this, we use higher derivatives.
You muat have coma across tho following series :
i) Exponential Series :
ii) Logarithmic Series :
Drawing Curves
Brook Taylor (1685-173 1) and Colin Maclaurin (1698 - 1746) were both d~sc~ples of Newton. Taylor first published his series in a paper In 1715. Maclaurin used Taylor's series . PS a fundamental tool in his work on calculus.
iii) Geometric Series :
ii) Binomial Series :
r (r - 1) (1 +x)'= 1 +rx+ ----- x2+ ............. providedIxI< 1
1.2
We observe that each of them is of the form
.......... ............. .......... f(x) = a, + a, x + a2x2 + a3x3 + + anxl1 + ........., where a,, a,, a2, all, are some constants.
We ask ourselves the questions. Is there anything else common to these four examples? Is it possible to express a,, a,, ......., a,, ........ in terms of the function f? Our answer is : Yes. In all these examples,
f'"' (0) a n = -
n!
In other words, the series is of the form
f ' (0) f'2' (0) f(x) = f(0) + - f'"' (0) ........ x+ - x2+ + - xn + ..........
l! 2! n!
We shall prove this for the above four instances, in the examples worked out below. .This expansion is called Taylor's series for f around zero. It is also known as Maclaurin's series for f. The name "Taylor's Series for f around zero" suggests that there npy be a Taylor's series for f around x,, (x,, # 0). But in this course we shall restrict ourselves only to the series around zero. This series expansion makes sense only when f is infinitely differentiable at zero. It is valid for many important functions (though not for all functions). You will learn mo,re about the validity of these series in the course on real analysis. In this section, you should train yourselves to write down Maclaurin's series for many functions.
We have said above that the function f should be infinitely differentiable, that is, it should have derivatives of all orders. How do we check this condition? For some functions it is not difficult. For example, we have
'cos hx, If n is a w n rlln hx. if n i u odd
In these cases, we can say that the derivatives of all orders exist for all values of x and we can Higher Order Derivatives
confidently use Taylor's series.
Example 10 Let us verify that the k n o w series expansion of ex is the same as its Maclaurin's series.
Maclaurin's series for ex is
Where f(x) = ex Now, f(0) = e0 = 1 Also, f (x) = ex :. f (0) = 1.
&fact, we know that e") (x) =ex for all natural numbers n, which means fit') (0) = e0 = 1 for all natural numbers n. Substituting these values of f(O), f (0), .......... en)(0) inMaciaurinls series we get
which is the known expansion for ex.
Example 11 Obtain Taylor's series for In (1 + x) around zero. Let fix) = ln (1 + x) Then we have already seen 1n Example 5, that
(n - I ) ! 1)ll-l -
(1 + x)"
Therefore, en)(0) = (- 1)"' (n- l)!
f'"' (0) - (-l)"-l . . . . n! n
Therefore Taylor's series around zero is
( -1) - (-1)2-' 7 (-l)"-I .... ..... ln (x) = --- X + ------ x- + +--- xI1 + 1 2 n
Hence, ~ac l iur in ' s series for ln (1 + x) is
We note that this is the same as the already known logarithmic series.
Example 12 Next, let us write down Maclaurin's series (or Taylor's series around zero) for ll(1 -x).
1 f (x) = ---- ; f (0) = 1 (1 - x ) ~
2 fi2) (x) = O] ; e2)(0)=2
We can prove by induction that n! f'"' (0) en) ( 4 =
(1 - x) n+l and therefore en) (0) = n! and hence - = 1.
n! Therefore Maclaurin's series is
Note that this agrees with what we already know, namely that the sum of the geometric series 1+x+x2+ ......... +xn+ ......... is 1/(1-x).
Example 13 Suppoee we want to wtite down Taylor's series for (1 + x)' around zero, where r is a fixed real number.
Let f(x) = (1 + x)'. Then f(0) .= 1 i ' (x)=r( l +x)'-' ; f (O)=t e2J(x)=r(r- 1)(1 +x)'-~ :fi2)(0)=r(r- 1 )
Drrwlng Curves We can prove by induction that fln)(x)=r(r- 1) ......( r-n+ 1)(1 +x)+"
....... fin)(0)=r(r- 1) (r-n+ 1)
Therefore Taylor's series around zero is
Note : This is the same as the binomial series that we already know. This expansion is valid only when I x I < 1. The reason for this will be clear when you study the course "Real Analysis". When r = -1, this binomial series becomes
Note that Example 12 follows from this on replacing x by - x throughout.
Further, we note that if r is a natural number than the series terminates after finite number of terms.
So far we have seen that the four known series occur as Taylor's series. In the next two examples, we find that we can write down similar series even for hnctions like sin x and cos x.
Example 14 Let us write down Maclaurin's series for the h c t i o n sin x.
Let f(x) = sin x. Then we have already seen in E 7)
0 if n is even
1 if n is of the form (4m + 1)
-1 otherwise, i.e., n is of the form (4m + 3)
nn fin) (0) = sin -
2
We see that, as n varies over 0, 1,2,3,4,5,6,7 ............ fen) (0) takes the values 0,1,0, -1,0,1,0, - 1, ......... Therefore Maclaurin's series for sin x is
Example 15 To fmd Taylor's series for cos 2x around zero, let us write f(x) = cos 2x. We have already seen in Example 6 that
nx fin) (0) = 2" cos -
2
Therefore, Taylor's series around zero is
Example 16 Suppow we want to a) write down the flnt four term of M.chrin'a d e a for tm x. b) write down the fint thm non m o tennr of thh nrioa.
Therefore the first four t e r n of Maclaurin's series for tan x are given by
1 0 2 2 3 0,-x,-x :--x , I ! 2! 3!
Hence Maclaurin's series for tan x is
x x+-+ .......
3
Now, we want the next non-zero term.
I We have f(*) (x) = 16 sec4 x tan x + 8 sec2 x tan3 x and 64) (0) = 0. .: tan (0) = 0.
~ e x t f ( ~ ) ( ~ ) = 1 6 s e c ~ x + 6 4 s e c ~ x t a n ~ x + 2 4 s e c ~ x t a n ~ x + 16sec2xtan4x
j This mans fiq (0) = 16.
1 Thus we have
I Example 17 If in Maclaurin's series for sin b , the coefficient of x3 is given to be - 6 k, let us
L find all possible values of k.
Maclaurin series for sin b is given by
kx k3x3 ....... --- + since qx) = k cos b l! 3!
f12) (x) = - sin b f13) (x) = - k3 cos b , where qx) = sin kx
The coefficient of x3 is - (k3/6) Therefore - (k3/6) = - 6k This gives the equation k (k2 - 36) = 0
: The roots are k = 0,6 or - 6.
Thus, 0 ,6 and - 6 are all the possible values of k such that the coefficient of x3 in Maclaurin's series for sin b is - 6k.
E E 19) Wri@ down Maclaurin's series for the following : 1
a) (l+x)z b) (~-2) '+1 c) COSX
d) l / ( l -2~)
Higher Order Derivatives
Drawing Curves
I
E E20) Write down the first three non-zero terms in Maclaurin's series of the following : I
a) sin3 x b) h(1-X) I
1 j
E E21) Find the coefficient of x9 in Maclaurin's series for the function. a) cos 2x
b) sin (X + :)
E E22) If Maclaurin's series for sin x is differentiated term by term, do you get Maclaurin's series Higher Order Derivatives
for cos x ?
E E 23) If Maclaurin's series of ex is differentiated tern by term, we get the same series again. Prove this.
E E 24) Consider the function y = a + tan-' bx where a and b are fixed real numbers. We are given t
that its Taylor's series arond zero is 2 + 3x - 9 x3 + ..... 1 Find the values of a and b.
Drawing Chrves E E25) Find the coefficient of x3 in Taylor's series around zero for the function sin-' x.
Some General Remarks on Taylor's and Maclaurin's Series
Though we have obtained infinite series for many functions, it is necessary to give a note of caution. These Infinite series need not be valid for all values of x, and as such, these have to be used with care. In the course on real analysis, you will be able to study the conditions under which these series are valid.
k # 0, since
k = O * O = q J ? ; ,
which is impossible.
1.6 SUMMARY
In this unit, we have
1) introduced higher order derivatives,
2) derived a formula (Leibniz's Theorem) for the nIh derivative of a product of two functions. (uv), = C(n, 0) u, v + C(n, 1) u,, v, + C(n, 2) u,, v2 + ........ + C(n, n) u v,.
3) written Taylor's series around zeroiMaclaurin's series of a number of functions by using the formula
1.7 SOLUTIONS AND ANSWERS
E l ) a) 6x E2) a) 11 JZ
E4) a) f(x) = sin kx * f12) (x) = - k2 sin kx * c2) (d6) = - k2 sin knl6
Now,-k2sinkd6=2&*sinkd6=-2&lk2
Since- 1 ~ s i n k ~ l 6 < O , - n < k x 1 6 < 0 *k=-1 or.-2or-3or-4or-5
Out of these, k = - 2 is the value which satisfies sin k A 16 = - 21 & lk2
E6) a) 3!an (ax + b)3-n, n I 3 ((3 - n)!
Higher Order 1)crivatives
m!an b, ((m - n)!
(ax + b)m-n, if n I m
c) ex d) kn ekx
i cos x if n =4k + 1 E7) f (x )=s inx~f (x )=cosx , f ' (x )~ -s inx , - s i n x i f n = 4 k + 2
f(3) (x) = - cos x, f(4) (x) = sin x and so on. So Our guess is that f (") (x) - - c o s x i f n = 4 k + 3 Now use the principle of mathematical induction to prove that f tn) (x) = sin Isin x if n = 4k
( X + y) as in Example 6.
E10) y =cosx*y,=-sinx,y2=-cosx,y3=sinx,y,=cosxandsoon. yn = cos (x + nd2)
3 yn+, = - sin (x + nd2) 3 yn2 + yn+,2 = cos2 (X + nd2) + sin2 (x + nd2) = 1.
E l l ) (uv),=u,v+5u4vl + 10u3v2+ 10u,v3+5ulv4+vS
E 12) (u v), = u, v + u v, which is the pribduct rule of differentiation.
d3(sin2 X) d3 = 7 (sinxsinx)=-cosxsinx-3 sinxcosx
dx3 dx
d - (sin2x) = 2 sin x cos x dx
(-I)"+, x3 E 15) x n [(n-l)!-3C(n, l)(n-2)!+6C(n,2)(n-3)!
- 6C (n, 3) (n - 4)!]
1 E21) a) 0 b, 9'JZ
I E 22) Yes
E24) We take that tan-' bx always takes values between --x/2 and d 2 . Then, a = 2, ,
b = 3
