$$y=(\mathrm{sin}x{)}^{\sqrt{x}}.$$

trapskrumcu
2022-09-26
Answered

Chain rule for the derivative of a composite function

$$y=(\mathrm{sin}x{)}^{\sqrt{x}}.$$

$$y=(\mathrm{sin}x{)}^{\sqrt{x}}.$$

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Zariah Fletcher

Answered 2022-09-27
Author has **8** answers

Use logarithmic differentiation here:

$$\begin{array}{rl}y& =(\mathrm{sin}(x){)}^{\sqrt{x}}\\ \mathrm{ln}(y)& =\sqrt{x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{ln}(\mathrm{sin}(x))\\ \frac{1}{y}\phantom{\rule{thinmathspace}{0ex}}\frac{dy}{dx}& =\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}{x}^{-1/2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{ln}(\mathrm{sin}(x))+\sqrt{x}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{cos}(x)}{\mathrm{sin}(x)}.\end{array}$$

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 ${x}^{x}$.

$$\begin{array}{rl}y& =(\mathrm{sin}(x){)}^{\sqrt{x}}\\ \mathrm{ln}(y)& =\sqrt{x}\phantom{\rule{thinmathspace}{0ex}}\mathrm{ln}(\mathrm{sin}(x))\\ \frac{1}{y}\phantom{\rule{thinmathspace}{0ex}}\frac{dy}{dx}& =\frac{1}{2}\phantom{\rule{thinmathspace}{0ex}}{x}^{-1/2}\phantom{\rule{thinmathspace}{0ex}}\mathrm{ln}(\mathrm{sin}(x))+\sqrt{x}\phantom{\rule{thinmathspace}{0ex}}\frac{\mathrm{cos}(x)}{\mathrm{sin}(x)}.\end{array}$$

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 ${x}^{x}$.

sombereki51

Answered 2022-09-28
Author has **3** answers

It's easier if you write

$$y={e}^{\sqrt{x}\mathrm{ln}(\mathrm{sin}(x))}$$

Then

$${y}^{\prime}={e}^{\sqrt{x}\mathrm{ln}(\mathrm{sin}(x))}(\sqrt{x}\mathrm{ln}(\mathrm{sin}(x)){)}^{\prime}=(\mathrm{sin}(x){)}^{\sqrt{x}}(\sqrt{x}\mathrm{ln}(\mathrm{sin}(x)){)}^{\prime}$$

Now you only have to differentiate $\sqrt{x}\mathrm{ln}(\mathrm{sin}(x))$, in which case you need the product rule and the chain rule.

$$y={e}^{\sqrt{x}\mathrm{ln}(\mathrm{sin}(x))}$$

Then

$${y}^{\prime}={e}^{\sqrt{x}\mathrm{ln}(\mathrm{sin}(x))}(\sqrt{x}\mathrm{ln}(\mathrm{sin}(x)){)}^{\prime}=(\mathrm{sin}(x){)}^{\sqrt{x}}(\sqrt{x}\mathrm{ln}(\mathrm{sin}(x)){)}^{\prime}$$

Now you only have to differentiate $\sqrt{x}\mathrm{ln}(\mathrm{sin}(x))$, in which case you need the product rule and the chain rule.

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