What is the derivative of $y = {(sin x)}^{x}$ $y=(sinx)^x$ ?

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What is the derivative of $y = {(sin x)}^{x}$ $y=(sinx)^x$ ?

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Answer 1 · 2015-07-27T15:50:45

$\frac{d y}{d x} = (ln (sin x) + x cot x) {(sin x)}^{x}$ $dy/dx = (ln(sinx)+xcotx)(sinx)^x$

Explanation:

Use logarithmic differentiation.

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

$ln y = ln ({(sin x)}^{x}) = x ln (sin x)$ $lny = ln((sinx)^x) = xln(sinx)$ (Use properties of $ln$ $ln$ )

Differentiate implicitely: (Use the product rule and the chain ruel)

$\frac{1}{y} \frac{d y}{d x} = 1 ln (sin x) + x [\frac{1}{sin x} cos x]$ $1/y dy/dx = 1ln(sinx) + x [1/sinx cosx]$

So, we have:

$\frac{1}{y} \frac{d y}{d x} = ln (sin x) + x cot x$ $1/y dy/dx = ln(sinx) + x cotx$

Solve for $\frac{d y}{d x}$ $dy/dx$ by multiplying by $y = {(sin x)}^{x}$ $y = (sinx)^x$ ,

$\frac{d y}{d x} = (ln (sin x) + x cot x) {(sin x)}^{x}$ $dy/dx = (ln(sinx)+xcotx)(sinx)^x$

Answer 2 · 2015-07-27T15:52:01

$\frac{d}{d x} {(sin x)}^{x} = (ln (sin x) + x cot x) {(sin x)}^{x}$ $d/dx(sinx)^x=(ln(sinx)+xcotx)(sinx)^x$

Explanation:

The easiest way to see this is using:

${(sin x)}^{x} = e^{ln ({(sin x)}^{x})} = e^{x ln (sin x)}$ $(sinx)^x=e^(ln((sinx)^x))=e^(xln(sinx))$

Taking the derivative of this gives:

$\frac{d}{d x} {(sin x)}^{x} = (\frac{d}{d x} x ln (sin x)) e^{x ln (sin x)}$ $d/dx(sinx)^x=(d/dxxln(sinx))e^(xln(sinx))$

$= (ln (sin x) + x \frac{d}{d x} (ln (sin x))) {(sin x)}^{x}$ $=(ln(sinx)+xd/dx(ln(sinx)))(sinx)^x$

$= (ln (sin x) + x \frac{\frac{d}{d x} sin x}{sin x}) {(sin x)}^{x}$ $=(ln(sinx)+x(d/dxsinx)/sinx)(sinx)^x$

$= (ln (sin x) + x \frac{cos x}{sin x}) {(sin x)}^{x}$ $=(ln(sinx)+xcosx/sinx)(sinx)^x$

$= (ln (sin x) + x cot x) {(sin x)}^{x}$ $=(ln(sinx)+xcotx)(sinx)^x$

Now we must note that if ${(sin x)}^{x} = 0$ $(sinx)^x=0$ , $ln ({(sin x)}^{x})$ $ln((sinx)^x)$ is undefined.

However, when we analyse the behaviour of the function around the $x$ $x$ 's for which this holds, we find that the function behaves well enough for this to work, because, if:

${(sin x)}^{x}$ $(sinx)^x$ approaches 0

then:

$ln ({(sin x)}^{x})$ $ln((sinx)^x)$ will approach $- \infty$ $-oo$

so:

$e^{ln ({(sin x)}^{x})}$ $e^(ln((sinx)^x))$ will approach 0 as well

Furthermore, we note that if $sin x < 0$ $sinx<0$ , $ln ({(sin x)}^{x})$ $ln((sinx)^x)$ will be a complex number; however, all the algebra and calculus that we have used work in the complex plane as well, so this is not a problem.

Answer 3 · 2017-08-14T17:20:25

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Explanation:

$d/dx [f(x)^g(x)] = [g(x)/f(x)f'(x) + g'(x)ln(f(x))][f(x)^g(x)]$

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