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Differential calculus (exercises with detailed solutions)

1. Using the definition, compute the derivative at x = 0 of the following functions:

a) 2x 5 b) x 3x 4 c)

x + 1 d) x sin x.

2. Find the tangent line at x = 1 of f(x) =x

x 2 .

3. Compute the derivatives of the following functions

a(x) = 2x3 9x + 7 cos x

b(x) = x sin x + cos x

c(x) =x2 4x2 + 4

d(x) =1 tan x1 + tan x

e(x) = x2 log x + 3x

f(x) =log x 1log x + 1

g(x) = x2 + 15

h(x) = log cos x + x tan x

i(x) =

4x2 3

l(x) = sin log

x

m(x) =5

sin3x

n(x) = arctan(1 x)

o(x) = x arcsin x +

1 x2

p(x) = log3x

.

4. Discuss the differentiability at x = 0 of the following functions

f(x) = sin |x|; g(x) = cos

|x|.

5. Let

f(x) =

(x )2 2, x 0 sin x, x < 0

.

Find , R such that f is continuous and differentiable on R.6. Discuss the differentiability of f(x) = |x2 1|.

7. Let f(x) = |x|x and g(x) = e1/x2 .Verify that is possible to extend f and g to continuous functions on R. Are the extended functionsdifferentiable at x = 0?

8. Write the equation of the tangent line to the graph of the function

f(x) =x + 2

x2 1 log(2x 3) .

at x = 2.

9. Study the monotonicity of f(x) = 3x + sin x.

10. Find local maxima and minima of the function f(x) = ex

+ 3ex

.

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Solutions

1. a) limx0

[2x 5 (5)]/x = 2; b) 1/16; c) limx0

[

1 + x 1]/x = limx0

[(1+ x) 1]/ x

1 + x + 1 = 1/2;d) 0.

2. The equation of the tangent line is y = f(1) + f(1)(x 1). Hence we compute the derivative of f atx0 = 1. Since f

(x) = 2(x2)2 , f(1) = 2, and the tangent line is y(x) = 1 2(x 1).

3. a) 6x2 9 7sin x, b) x cos x, c) 16x(x2 + 4)2

, d) 2(sin x + cos x)2

, e) 2x log x + x + 3,

f)2

x(log x + 1)2g) 10x(x2 + 1)4 h)

x

cos2 xi)

4x4x2 3 l)

1

2xcos log

x

m)3cos3x

55

sin4 3xn) 1

x2 2x + 2 o) arcsin x p) log(log 3)

log3x

.

4. At x = 0 f does not exists. Indeed the right an left derivatives of f at x = 0 are

lim x x+ sin x 0x

= 1 and lim x x sin(x) 0x

= 1.

The right derivative of g at 0 is 1/2, the left one is 1/2, hence g is not differentiable at x = 0.5. f is continuous and differentiable if x = 0; f is continuous at x = 0, if = 2.

The derivative of f is f(x) = 2(x ) if x > 0 and f(x) = cos x if x < 0, We now compute the limitof f(x) as x 0: if such limit exists and assumes a value R then f is differentiable at x = 0 andf(0) = . We have lim

x0+f(x) = 2 and lim

x0f(x) = , hence f is differentiable if = 22 and

=

2 or = 2

2 and = 2.6. f is differentiable on R \ {1, 1}. At x = 1, f f is not differentiable (the right derivative is 2, the left

one

2); at x = 1, f is not differentiable (the right derivative is 2, the left one

2).

If x = 1, f is x2 1 or 1 x2, hence it is differentiable and its derivative is respectively 2x and 2x.7. We can extend f to a continuous function at x = 0. Indeed, since |x|x := ex log |x|, we have

limx0

|x|x = limx0

ex log |x| = 1.

Let

f(x) :=

|x|x, if x = 01, if x = 0

;

then

limx0

f(x) f(0)x

0

= limx0

|x|x 1x

= limx0

ex log |x| 1

x log

|x

|

log |x| = .

f is not differentiable at 0.

The function g can be extended to a continuous function at x = 0. Indeed

limx0

e1

x2 = 0.

Let

g(x) :=

e

1

x2 , if x = 0

0, if x = 0;

Hence

limx0

g(x) g(0)x 0 = limx0

e1

x2 0x

= limt

t et2

= 0.

The extention g is then differentiable at 0, and g

(0) = 0 .

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8. We remark that f is differentiable at x = 2 and that, whenever x > 3/2,

f(x) =1

x2

1 2x(x + 2)

(x2

1)2

2

2x 3.

In particular, f(2) = 31/9. Since f(2) = 4/3, the tangent line is y = 31/9 x + 74/9.9. We compute f(x) = 3 + cos x and we impose that f(x) 0. Since this inequality is satisfied for every

x R, f f increases on R.10. dom(f) = R and f is differentiable on R. Since f(x) = ex 3ex, we have that f(x) > 0 if e2x > 3,

that is if x > 12

log 3; on the other hand f(x) < 0 if x < 12

log 3. Hence f has minimum at x = 12

log 3.

11. f(x) = 3 cos x 3, hence f(0) = 0 and f(x) < 0 for every x sufficiently close to 0 (except 0). x0 = 0 isa point of inflection. For the function g, x0 = 0 is a point of inflection.

12. f is continuous and strictly increasing, hence invertible, on R. Furthermore, since f never vanishes onR

, the inverse function f

1

is differentiable for every x dom(f1

) =R

. For every x R

we havef1

(x) =

1

f(f1(x))=

1

7 (f1(x))6 + 1.

Since f(0) = 0, f1(0) = 0, hence

f1

(0) = 1.

Since f(1) = 2 , f1(2) = 1 , hence

f1

(2) = 1/8.

13. Since dom(f) = R \ {(3 2, 3 + 2)}, we can look for oblique asymptotes at + and at .We compute

limx

f(x)

x= lim

x

|x|x

1 6

x+

7

x2= 1.

Furthermore

limx+ (f(x) 1 x) = limx+ 6x + 7

x2 6x + 7 + x = 3and

limx

(f(x) + 1 x) = limx

6x + 7x2 6x + 7 x = 3.

Hence f has an oblique asymptote at +, y = x 3, and an oblique asymptote at +, y = x + 3.14. In the first case we impose that x2 = 2/15, we obtain x = 2/15.

In the second one we impose 15x2 = 1, which does not admit any solution.15. The slopes of the tangent lines to the graphs of f and g are respectively m(x) = 3x2 on R and n(x) =

1/(3x2) on R \ {0}. Since for every x = 0 we have m(x) n(x) = 1, the tangent lines to f and g areorthogonal.

16. dom(f) = (2, +) and limx f(x) = . Furthermore f is strictly increasing on dom(f). Hence fhas a unique zero at x0 = 1.

17. f is continuous on [0, 1] and differentiable on (0, 1), the assumptions of the Lagranges Theorem aresatisfied. Furthermore we have = 1/3.

18. If x = 0 the function is differentiable. At x = 0 we compute the limit:

limx0

2x2 + x2 sin 1xx

= limx0

2x + x sin

1

x

= 0,

hence f is differentiable on R.

f(x) = 4x + 2 sin(1/x) cos(1/x) when x = 0 and f(0) = 0. Since limx0 f(x) does not exists, f isnot continuous on R.

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19. f must be continuous on [0, 1] and differentiable on (0, 1). f satisfies these assumptions if x2 = , forevery x [0, 1].If < 0, such condition is always satisfied.

If = 0, is not satisfied (at x = 0 the denominator vanishes).If > 0, we need to impose x = for every x [0, 1], that is x = for every x [0, 1]. This isequivalent to ask that

/ [0, 1] that is / [0, 1]. We conclude that the assumptions of the Lagranges

Theorem are satisfied for every / [0, 1].20. If a = 0 f is trivially convex; more generally, f(x) = 2 a cos x is positive if a cos x 2. If a > 0, this

inequality is trivially satisfied for every x such that cos x < 0. If cos x 0, we obtain 0 a cos x a 2,hence we need 0 a 2. If a < 0, the inequality a cos x 2 is satisfied for every a if cos x > 0. If cos x 0, then the inequality is |a cos x| 2 and we deduce |a| 2.Conclusions: f is convex on R if |a| 2.

21. Let f(x) := (1 + x), x > 1. We have f(x) = (1 + x)1 and f(x) = ( 1)(1 + x)2. Since fis positive, f is convex, hence

f(x) f(0) + f(0)(x 0) = 1 + x.

We can also argue in a different way defining g(x) := (1 + x) 1 x, as x > 1. We remark thatg(0) = 0, g(x) > 0 when x > 0 and g(x) < 0 when 1 < x < 0. Then g(x) 0 for every x > 1.

22. Let f(x) := sin x x + x36

, x 0. We have f(x) = cos x 1 + x22

and f(x) = sin x + x. We know thatsin x x for every x 0, hence f(x) 0 for every x 0 and f is convex. We then have the estimatef(x) f(0) + f(0)(x 0) = 0 for every x 0.

23. a) Let f(x) = x3 + 3x2 + 5x + 3; f as x and f is strictly increasing: the equation hasexactly one real solution.b) Let g(x) = x4 + 4x + 12; g

+

as x

; g has a minimum at x =

1, g(

1) = 9. Hence the

equation g(x) = 0 has no real solutions.c) Let h(x) = 3x55x3+1; h as x . h(x) = 15x2(x21), hence h is strictly increasing whenx < 1 or x > 1, decreasing in (1, 1); x = 0 is a point of inflection. Since h(1) = 3 and h(1) = 1,h(x) = 0 has 3 real solutions (one smaller then 1, one between 1 and 1, one greater then 1).

24. True; indeed if f(x) = x5 + x we have limx+ = + and limx = . Hence the equation has atleast one solution. The solution is unique since f is strictly increasing

25. We study f(x) = x log x, defined when x > 0. We have limx0+ f = 0 and limx+ f = +. Further-more f(x) = log x + 1, hence f increases when x > 1/e. f has a minimum at 1/e, and f(1/e) = 1/e.We then obtain: if < 1/e we have no solutions; if = 1/e we have one solution; if 1/e < < 0 wehave two solutions; if 0 we have one solution.

26. We study the function f(x) = xe

x

, defined for every x R

. We have limx = 0 and limx+ =+. Furthermore, f(x) = ex(x + 1), hence f increases if x > 1. f has a minimum if x = 1 andf(1) = 1/e. We then obtain: if < 1/e we have no solutions; if = 1/e we have one solution; if1/e < < 0 we have two solutions; if 0 we have one solution.

27. We study f(x) = x2 arctan x, defined for every x R. We have limx = and limx+ = +.Furthermore, f(0) = 0 and f(x) = 2x arctan x + x2/(1 + x2). Since x arctan x 0 (and vanishes if andonly if x = 0) and 1/(x2 + 1) 1 (the equality holds if and only if x = 0), we conclude that f(x) > 0 forevery x R, x = 0. Hence f is strictly increasing. The equation admits a unique solution for every valueof a.

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