MATH 140 Lecture 33: Logarithmic Differentiation

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G( x)= x a f (t )dt , a x b g" (x)=f (x ) if f is continuous on [a, b]. G( x)= h( x ) o f (t )dt g"( x)=f (k (x))k" ( x) f (h( x))h"( x) Quiz 9, problem 2 x ( 8) sin(t 2)dt=(sin( x42))(4 x3)=4 x3 sin x 4 d dx . 0 x2 4 0 for [2,3]; x2 4 0 for[0,2] o. T 5 t 3+1dt: let u=t 3+1 u 1=t3 du=3t2 dt o. Integration involving logarithms: let r 1, then: xr dx= 1 r+1 xr +1+c. 2)+ 3/rt ow m2le on[a ,b],thenlet :s that :c u: if r= 1 , then we have 1 x dx , 1 x is continuouson(0, ) x 1, refine: g( x)= t. 1 dt for x>1 g"( x)= 1 x. 1 dt=0 and we know that ln 1=0 g ( x)=ln x for all x 1: thus: x 1 ln x= t.