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The Chain Rule If f(x) = g [ h(x) ], then f’(x) = g’ [ h(x)] h’(x) = The derivative of the outside times the derivative of the inside Ex: Find the derivatives of these functions 1) ex 2) e-x 3) e3x
4) (3x – 4) 5) (3x – 4)2
¥ky=e×
( e- 5'
= E×
( x )'
= e-×
C-D= - e-×
⇐ (e3D=e3×¥GD= 3 @
3×
¥ (3×-4)=3.
¥ ( 3×-45= 2 (3×-4)
"
# (3*4)= 2 (3×-4) (3)= 6 (3×-4)=18×-24
galionAt (3×-45off( g)Examination\ 6( 3×-4
#uk¥
Integration by u Substitution Ex: Integrate 1) !"#$ 2) 2!&"#$ 3) !&"#$
= ex + C
I ( ? )=2e2×
£,G2D=2e "
S2e2×d×=e2×+C
¥ ( ? )=e"
Let u=2×
onmgljwiabk '
itSe2×dx= Seodx
U=2×
du=2dx
de = dx
= Seo¥=±eu+C=Ie2×tC
4) 12(4$ − 2)$ 5) (4$ − 2)$ 6) 4$ − 2#$
u . - 4×-2
du=4dx or d¥=d×
Siz ( 42¥or
3C
= S3u2du or 3µdu /= u3+C or 3 . § + ( = ✓ + (
=(4× . 2)3+ (
U= 4×-2 .
du=4dx
dgu = dx
Scu )2¥= ytfidu:÷¥f⇒t¥¥+c
= | ( 4×-2 )÷dx0=4×-2du=4dx
d¥=dx= Sws±d÷=tfu÷du
=¥¥÷+c=¥÷E+c=f(4*ykt( = 's fyxt )3tC
why use substitution ?
) x3dx = ¥ + C
) ( x+D3dxU= xtl
du = dx to
= f ( updo= yI+c= k¥4 't
¥ @¥'+D= 4k¥37)= ( xti )3
/§Cx+D3dx
The limits¥44u=x+i
u= XH×=( : U=Z
du= DX ×=3 : 0=4
= { Vdu or Mdu=¥3i :÷¥y
= k¥4 ] ? :=¥ . ¥= '4¥ . c¥=2¥,¥
= 64-4
= 64-4=60= 60