Finding the derivative via implicit differentiation of an exponent?

Question

Am I doing this one correct. Wolfram Alpha says otherwise but I'm not sure where I go wrong:

$$y = x^{\sin (x)}$$

Wolfram alpha says the derivative is:

$$y'(x) = x^{(\sin(x) - 1)} \cdot (\sin(x) + x \cdot \log(x) \cdot \cos(x))$$

But here are my calcs:

$$\ln y = \sin(x) \cdot \ln(x)$$

$$\frac{1}{y} \cdot \frac{dy}{dx} = \frac{\sin(x)}{x} + \ln(x)\cos(x)$$

$$\frac{dy}{dx} = x^{\sin(x)} \cdot \left(\frac{\sin(x)}{x} + \ln(x)\cos(x)\right)$$

Answer 1

You have the same answer since

$$x^{\sin(x)}\left(\frac{\sin(x)}{x}+\cos(x)\log(x)\right)=x^{\sin(x)-1}\left(\sin(x)+x\log(x)\cos(x)\right)$$ multiply the bracket by $x$ and $x^{\sin(x)}$ by $\frac1x$

Answer 2

$$y=x^{\sin(x)}$$ Take the logarithm of both sides: $$\log(y)=\sin(x)\log(x)$$ Now differentiate both sides with respect to $x$: $$\frac{y'}{y}=\frac{\sin(x)}{x}+\cos(x)\log(x)$$ $$y'=y*\left(\frac{\sin(x)}{x}+\cos(x)\log(x)\right)$$ $$y'=x^{\sin(x)}\left(\frac{\sin(x)}{x}+\cos(x)\log(x)\right)$$ WolframAlpha's answer: $$y'=x^{\sin(x)-1}\left(\sin(x)+x\log(x)\cos(x)\right)$$ $$y'=\frac{x^{\sin(x)}}{x^1}\left(\sin(x)+x\log(x)\cos(x)\right)$$ $$y'=x^{\sin(x)}\left(\frac{\sin(x)}{x}+\frac{\log(x)\cos(x)x}{x}\right)$$ So both of the answers are the same.

Answer 3

Note that the two expression are equal, indeed

$$x^{(\sin x - 1)}=x^{\sin x}\cdot x^{-1}$$

and thus

$$y'(x) = x^{(\sin x - 1)} \cdot (\sin x + x \cdot \log x \cdot \cos x )= x^{\sin x } \cdot\frac1x (\sin x + x \cdot \log x \cdot \cos x)$$

Answer 4

Use the general formula

$$\frac{d}{dx} f(x)^{g(x)} = f(x)^{g(x)}\left(g'(x)\ln(f(x)) + g(x)\frac{f'(x)}{f(x)}\right)$$