What is the derivative of x^sin(x)?

Dec 20, 2016

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

Explanation:

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

Take the natural logarithm of both sides.

$\ln y = \ln \left({x}^{\sin} x\right)$

Use laws of logarithms to simplify.

$\ln y = \sin x \ln x$

Use the product rule and implicit differentiation to differentiate.

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

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

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{\cos x \ln x + \sin \frac{x}{x}}{\frac{1}{y}}$

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

Hopefully this helps!

Dec 20, 2016

$\frac{d}{\mathrm{dx}} {x}^{\sin x} = {x}^{\sin x} \left[\cos x \cdot \ln x + \frac{\sin x}{x}\right]$.

Explanation:

When we have a function of $x$ like $y = {x}^{\sin} x$, where a single term contains $x$ in both its base and its power, perhaps the easiest way to find the function's derivative is to first take the (natural) logarithm of both sides:

$\ln y = \ln \left({x}^{\sin x}\right)$
$\textcolor{w h i t e}{\ln y} = \sin x \cdot \ln x$

This places all the $x$'s on the same "level". Then, take the derivative of both sides with respect to $x$:

$\implies \frac{d}{\mathrm{dx}} \left(\ln y\right) = \frac{d}{\mathrm{dx}} \left(\sin x \cdot \ln x\right)$

Remembering that $y$ is a function of $x$, we get

$\implies \frac{1}{y} \cdot \frac{\mathrm{dy}}{\mathrm{dx}} = \cos x \cdot \ln x + \sin x \left(\frac{1}{x}\right)$
$\implies \textcolor{w h i t e}{\text{XXi}} \frac{\mathrm{dy}}{\mathrm{dx}} = y \left[\cos x \cdot \ln x + \frac{\sin x}{x}\right]$

Since we began with $y = {x}^{\sin x}$, we substitute this back in for $y$ to get

$\implies \textcolor{w h i t e}{\text{XXi}} \frac{\mathrm{dy}}{\mathrm{dx}} = {x}^{\sin x} \left[\cos x \cdot \ln x + \frac{\sin x}{x}\right]$.

Note:

When $f \left(x\right) = g {\left(x\right)}^{h \left(x\right)}$, you'll almost always see $g {\left(x\right)}^{h \left(x\right)}$ appear in the derivative of $f \left(x\right)$. If you don't, go back and double check your work to make sure things were done right.

Dec 20, 2016

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

Explanation:

Given: $y = {x}^{\sin \left(x\right)}$

Use logarithmic differentiation.

$\ln \left(y\right) = \ln \left({x}^{\sin \left(x\right)}\right)$

On the right side, use a property of logarithms, $\ln \left({a}^{b}\right) = \left(b\right) \ln \left(a\right)$:

$\ln \left(y\right) = \left(\sin \left(x\right)\right) \ln \left(x\right)$

Use implicit differentiation on the left side:

$\frac{\mathrm{dl} n \left(y\right)}{\mathrm{dx}} = \frac{1}{y} \frac{\mathrm{dy}}{\mathrm{dx}}$

Use the product rule on the right sides:

$\frac{d \left(u v\right)}{\mathrm{dx}} = \left(u '\right) \left(v\right) + \left(u\right) \left(v '\right)$

let $u = \sin \left(x\right)$, then $u ' = \cos \left(x\right) , v = \ln \left(x\right) , \mathmr{and} v ' = \frac{1}{x}$

Substituting into the product rule:

$\frac{d \left(\left(\sin \left(x\right)\right) \ln \left(x\right)\right)}{\mathrm{dx}} = \left(\cos \left(x\right)\right) \ln \left(x\right) + \frac{\sin \left(x\right)}{x}$

Put the equation back together:

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

Multiply both sides by y:

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

Substitute ${x}^{\sin \left(x\right)}$ for y:

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