Question

use logarithmic differentiation to find derivative: (1+sin(x))^(1/x)??

Answer 1

\[y=(1+\sin(x))^\frac{1}{x}\]
Take natural log of both sides
\[\ln(y)=\ln(1+\sin(x))^\frac{1}{x}\]
\[\ln(y)=\frac{1}{x} \ln(1+\sin(x))\]
now we take derivative of both sides
we need to recall the product rule and chain rule
\[[\ln(y)]'=[\frac{1}{x}]'(1+\sin(x))+\frac{1}{x}[1+\sin(x)]'\]
----
\[[\ln(y)]'=\frac{y'}{y}\]
\[[\frac{1}{x}]'=\frac{-1}{x^2}\]
\[[1+\sin(x)]'=0+\cos(x)=\cos(x)\]
--------------------------

Answer 2

opps i left my natural log off

Answer 3

\[[\ln(y)]'=[\frac{1}{x}]'\ln(1+\sin(x)+\frac{1}{x}[\ln(1+\sin(x)]'\]

Answer 4

\[[\ln(1+\sin(x)]'=\frac{[1+\sin(x)]'}{1+\sin(x)}=\frac{0+\cos(x)}{1+\sin(x)}=\frac{\cos(x)}{1+\sin(x)}\]

Answer 5

\[\frac{y'}{y}=\frac{-1}{x^2} \cdot \ln(1+\sin(x))+\frac{1}{x} \cdot \frac{\cos(x)}{1+\sin(x)}\]

Answer 6

multiply y on both sides to find y'

Answer 7

ok i got it you gave me the start i needed