Question

find dy/dx using logarithmic differentiation: y = x^cosx CLICK to see my work so far...

Answer 1

lny=lnx^cosx --> lny=cosxlnx --> (1/y)(dy/dx)=(lnx)(-sinx)

Answer 2

\[\ln y=\cos x~\ln x\]
\[\frac{ 1 }{ y }\frac{ dy }{ dx }=\cos x*\frac{ 1 }{ x }+\left( -\sin x \right)\ln x\]
\[\frac{ dy }{ dx }=?\]

Answer 3

what is (1/y)(dy/dx)? (dy/dyx)?

Answer 4

\[\frac{ dy }{ dx }=y \left( \frac{ \cos x }{ x }- \sin x~\ln x \right)\]
substitute the value of y

Answer 5

and that would be my final answer, because i cannot simplify anymore!

Answer 6

thank you so much!

Answer 7

@Amenah8 wait wait wait - did u substitute the formula for y back in?

Answer 8

when he said to substitute the value for y, did he mean to put in y for something else?

Answer 9

@surjithayer suggested that u should substitute y = x^cosx so that dy/dx is expressed only in x

Answer 10

so put into x^cosx for y?

Answer 11

(dy/dx) = (x^cosx))(cosx/x)-sinxlnx) ?

Answer 12

correct :) sorry about my late reply!

Answer 13

it's okay! that's my final answer, right?

Answer 14

well i THINK so...but dont kill me if im wrong! lol

Answer 15

haha thank you!

Answer 16

welcome