# Chain rule for the derivative of a composite functiony=(sinx)sqrt x.

Chain rule for the derivative of a composite function
$y=\left(\mathrm{sin}x{\right)}^{\sqrt{x}}.$
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Zariah Fletcher
Use logarithmic differentiation here:
$\begin{array}{rl}y& =\left(\mathrm{sin}\left(x\right){\right)}^{\sqrt{x}}\\ \mathrm{ln}\left(y\right)& =\sqrt{x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{ln}\left(\mathrm{sin}\left(x\right)\right)\\ \frac{1}{y}\phantom{\rule{thinmathspace}{0ex}}\frac{dy}{dx}& =\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}{x}^{-1/2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{ln}\left(\mathrm{sin}\left(x\right)\right)+\sqrt{x}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{cos}\left(x\right)}{\mathrm{sin}\left(x\right)}.\end{array}$
The problem with the straight-forward approach is that you don't have a power rule, nor do you have an exponent rule you can use. It's a bit analogous to differentiating ${x}^{x}$.
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sombereki51
It's easier if you write
$y={e}^{\sqrt{x}\mathrm{ln}\left(\mathrm{sin}\left(x\right)\right)}$
Then
${y}^{\prime }={e}^{\sqrt{x}\mathrm{ln}\left(\mathrm{sin}\left(x\right)\right)}\left(\sqrt{x}\mathrm{ln}\left(\mathrm{sin}\left(x\right)\right){\right)}^{\prime }=\left(\mathrm{sin}\left(x\right){\right)}^{\sqrt{x}}\left(\sqrt{x}\mathrm{ln}\left(\mathrm{sin}\left(x\right)\right){\right)}^{\prime }$
Now you only have to differentiate $\sqrt{x}\mathrm{ln}\left(\mathrm{sin}\left(x\right)\right)$, in which case you need the product rule and the chain rule.