Question

find the derivative of y=(sinx)^x

Answer 1

Oo this looks like a fun one :O

Answer 2

Maybe start by taking the natural log of each side,\[\large\rm \ln y=\ln\left[(\sin x)^x\right]\]Apply your log rule: \(\large\rm \log(a^b)=b\cdot\log(a)\)
giving us,\[\large\rm \ln y=x\cdot\ln\left[\sin x\right]\]Ok with those steps?

Answer 3

Our differentiation becomes doable from this point.

Answer 4

that's as far as I get

Answer 5

Differentiate with respect to x,\[\large\rm \color{royalblue}{\left(\ln y\right)'}=\color{royalblue}{\left(x\right)'}\ln\left[\sin x\right]+x\color{royalblue}{\left(\ln\left[\sin x\right]\right)'}\]Here is the "set up" for differentiating.

The things in blue are where we need to take a derivative.
I applied product rule to the right side of the equation.

Answer 6

Do you understand how to differentiate the `left side of the equation`?

Answer 7

so it would be ln(sinx)+x(1/x)?

Answer 8

On the `right side of the equation`? Hmm not quite.\[\large\rm \color{royalblue}{\left(\ln y\right)'}=\color{royalblue}{\left(x\right)'}\ln\left[\sin x\right]+x\color{royalblue}{\left(\ln\left[\sin x\right]\right)'}\]\[\large\rm \color{royalblue}{\left(\ln y\right)'}=\color{orangered}{\left(1\right)}\ln\left[\sin x\right]+x\color{royalblue}{\left(\ln\left[\sin x\right]\right)'}\]This first derivative looks ok.
Let's look closer at log(sinx) though

Answer 9

So you know that: \(\large\rm \frac{d}{dx}\ln(x)=\frac{1}{x}\)
Good.

But we need to generalize it,
so we can deal with what happens when there is more than just x inside the log.
\[\large\rm \frac{d}{dx}\ln(stuff)=\frac{1}{stuff}(stuff)'\]It's 1 over the stuff,
not just 1/x, but the entire stuff inside the log,
and then chain rule says to multiply by the derivative of that stuff.

Answer 10

\[\large\rm \frac{d}{dx}\ln(\sin x)=\frac{1}{\sin x}(\sin x)'\]
\[\large\rm \frac{d}{dx}\ln(\sin x)=\frac{1}{\sin x}(\cos x)\]ok with those steps?\[\large\rm \frac{d}{dx}\ln(x)=\frac{\cos x}{\sin x}\]

Answer 11

woops* that last line should be:\[\large\rm \frac{d}{dx}\ln(\sin x)=\frac{\cos x}{\sin x}\]

Answer 12

\[\large\rm \color{royalblue}{\left(\ln y\right)'}=\color{royalblue}{\left(x\right)'}\ln\left[\sin x\right]+x\color{royalblue}{\left(\ln\left[\sin x\right]\right)'}\]\[\large\rm \color{royalblue}{\left(\ln y\right)'}=\color{orangered}{\left(1\right)}\ln\left[\sin x\right]+x\color{royalblue}{\left(\ln\left[\sin x\right]\right)'}\]\[\large\rm \color{royalblue}{\left(\ln y\right)'}=\color{orangered}{\left(1\right)}\ln\left[\sin x\right]+x\color{orangered}{\left(\frac{\cos x}{\sin x}\right)}\]

Answer 13

But that is not your `answer`.
Differentiate the left side.
It contains more than just dy/dx.
There are a few more steps.