Question

DEFFERENTIATE X^sinx

Answer 1

Answer:

y = x^(sin(x))

y = e^(SIN(x)*LN(x))

y' = (COS(x)·LN(x) + SIN(x)/x)*e^(SIN(x)*LN(x))

y' = (COS(x)·LN(x) + SIN(x)/x)*x^(SIN(x))

Answer 2

Answer:

$\boxed{(x^{\sin x})' =x^{\sin x}\left(\cos x \log x + \dfrac{\sin x}{x}\right)}.$

Step-by-step explanation:

First, we should rewrite the expression as:

$x^{\sin x} = \exp[\log(x^{\sin x})] = \exp(\sin x \log x) = e^{\sin x \log x}.$

We now use the chain rule to get:

$(e^{\sin x \log x})' = e^{\sin x \log x}(\sin x \log x)' = x^{\sin x}(\sin x \log x)'.$

The derivative of the product is given by:

$(\sin x \log x)' = (\sin x)' \log x + \sin x (\log x)' = \cos x \log x + \dfrac{1}{x}\sin x.$

Substituiting, we get:

$x^{\sin x}(\sin x \log x)' = x^{\sin x}\left(\cos x \log x + \dfrac{\sin x}{x}\right).$

So the answer is:

$\boxed{(x^{\sin x})' =x^{\sin x}\left(\cos x \log x + \dfrac{\sin x}{x}\right)}.$