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