# Derivative of (sin(x))^x

1. Apr 5, 2004

### The_Brain

What are the steps used to obtain the derivative of (sin(x))^x? I know it's (sin(x))^x [xcot(x) + ln(sinx)] however I don't know how to get there.

2. Apr 6, 2004

### faust9

Ahhhh a "tough" question. Most texts will show the proof for $\frac{d}{du}a^u$ and $\frac{d}{du}u^a$ where a is a constant and u is a function. What you have is a combination of the two where the base is a function raised to the power of a function.

Lets look at a simple proof that should help you with the above.

$$y=U^V$$ where u and v are variables.

$$\ln{y}=\ln{U^V}$$ we'll do a little log magic.

$$\ln{y}=V\ln{U}$$ remember how we deal with powers within log function.

$$\frac{d}{dx}(\ln{y}=V\ln{U})$$ we'll do some implicit differentiation.

$$\frac{dy}{dx}\frac{1}{y} = V\frac{1}{U}\frac{dU}{dx} + \frac{dV}{dx}\ln {U}$$ remember your chain rule.

$$\frac{dy}{dx} =y (V \frac{1}{U} \frac{dU}{dx} + \frac{dV}{dx} \ln {U})$$ move our y over remember $y=U^V$.

$$y^\prime =U^V (V U^{-1} U^\prime + V^\prime \ln {U})$$ change our notation to make things look cleaner.

thus we have
$$y^\prime = V U^{V-1} U^\prime + U^VV^\prime \ln {U}$$

The above should give you the answer you are looking for. One recomendation if you don't do this already: for rules that require a mixture of prime and non prime functions I'd make a table like this:

given
$$y=x\sin{x}$$
$$\frac{d}{dx}(y=x\sin{x})$$ you need the chain rule here.

let $$U=x$$ and $$V=\sin{x}$$
thus:$$U^\prime=1$$ and $$V^\prime=\cos{x}$$

now just plug these into the chain rule $y^{\prime}=UV^{\prime}+U^{\prime}V$

Writing your information in a clean tabular format will mean you are less likely to make a mistake when plugging in the primes and non prime functions into the rule. Just a thought.

Good Luck

3. Apr 6, 2004

### The_Brain

Of course! Differentiation by logarithms. Thanks.