1. ## log deriv.

Use logarithmic differentiation to find the derivative of the function.
y = x^sinx

How do you do this? Help please.

2. Originally Posted by kwivo
Use logarithmic differentiation to find the derivative of the function.
y = x^sinx

How do you do this? Help please.
begin by taking the log of both sides.

$\displaystyle y = x^{\sin x}$

$\displaystyle \Rightarrow \ln y = \ln x^{\sin x}$

$\displaystyle \Rightarrow \ln y = \sin x \ln x$

differentiating implicitly with repsect to x, we get:

$\displaystyle \frac {y'}y = \cos x \ln x + \frac {\sin x}x$

$\displaystyle \Rightarrow y' = y \left( \cos x \ln x + \frac {\sin x}x\right) = x^{\sin x}\left( \cos x \ln x + \frac {\sin x}x\right)$

3. Hello, kwivo!

Use logarithmic differentiation to differentiate: .$\displaystyle y \:=\:x^{\sin x}$

We have: .$\displaystyle y \;=\;x^{\sin x}$

Take logs: .$\displaystyle \ln y \:=\:\ln\left(x^{\sin x}\right)\quad\Rightarrow\quad\ln y \:=\:\sin x\cdot\ln x$

Differentiate implicitly: .$\displaystyle \frac{1}{y}\cdot\frac{dy}{dx} \;=\;\sin x\cdot\frac{1}{x} + \cos x\cdot\ln x$

Multiply by $\displaystyle y\!:\;\;\frac{dy}{dx}\;=\;y\left[\frac{\sin x}{x} + \cos x\!\cdot\!\ln x\right]$

Replace $\displaystyle y\!:\;\;\frac{dy}{dx}\;=\;x^{\sin x}\left[\frac{\sin x}{x} + \cos x\!\cdot\!\ln x\right]$

4. Originally Posted by kwivo
Use logarithmic differentiation to find the derivative of the function.
y = x^sinx
I'll use the never well weighed trick $\displaystyle a=e^{\ln a},\,\forall a>0,$ then $\displaystyle y=e^{\sin x\cdot\ln x}$

Now it remains to compute $\displaystyle (\sin x\cdot\ln x)',$ and yields the desired.