# What is the derivative of f(theta)=arcsin(sqrt(sin(9theta)))?

Aug 29, 2015

$\frac{\mathrm{df}}{d \theta} = \frac{9}{2} \cdot \cos \frac{9 \theta}{\sqrt{\sin \left(9 \theta\right)} \cdot \sqrt{1 - \sin \left(9 \theta\right)}}$

#### Explanation:

You can differentiate this function by using implicit differentiation, provided that you don't know what the derivative of $\arcsin \left(x\right)$ is.

You will also need to use the chain rule and the fact that

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

$f \left(\theta\right) = \arcsin \left(\sqrt{\sin \left(9 \theta\right)}\right)$

is equivalent to

$\sin \left(f\right) = \sqrt{\sin \left(9 \theta\right)} \text{ } \textcolor{\mathmr{and} a n \ge}{\left(1\right)}$

Differentiate both sides with respect to $\theta$

$\frac{d}{d \theta} \left(\sin \left(f\right)\right) = \frac{d}{d \theta} \sqrt{\sin \left(9 \theta\right)}$

Focus on finding $\frac{d}{d \theta} \sqrt{\sin \left(9 \theta\right)}$ first. Use the chain rule twice to get

$\frac{d}{d \theta} \sqrt{\sin \left(9 \theta\right)} = \frac{1}{2} \sin {\left(9 \theta\right)}^{- \frac{1}{2}} \cdot \cos \left(9 \theta\right) \cdot 9$

$\frac{d}{d \theta} \sqrt{\sin \left(9 \theta\right)} = \frac{9}{2} \sin {\left(9 \theta\right)}^{- \frac{1}{2}} \cdot \cos \left(9 \theta\right)$

Take this back to your target derivative to get

$\cos \left(f\right) \cdot \frac{\mathrm{df}}{d \theta} = \frac{9}{2} \sin {\left(9 \theta\right)}^{- \frac{1}{2}} \cdot \cos \left(9 \theta\right)$

Isolate $\frac{\mathrm{df}}{d \theta}$ on one side

$\frac{\mathrm{df}}{d \theta} = \frac{9}{2} \cdot \frac{\sin {\left(9 \theta\right)}^{- \frac{1}{2}} \cdot \cos \left(9 \theta\right)}{\cos} \left(y\right)$

Use the trigonometric identity

$\textcolor{b l u e}{{\sin}^{2} x + {\cos}^{2} x = 1}$

to write $\cos x$ as a function of $\sin x$

${\cos}^{2} x = 1 - {\sin}^{2} x \implies \cos x = \sqrt{1 - {\sin}^{2} x}$

This will get you

$\frac{\mathrm{df}}{d \theta} = \frac{9}{2} \cdot \frac{\sin {\left(9 \theta\right)}^{- \frac{1}{2}} \cdot \cos \left(9 \theta\right)}{\sqrt{1 - {\sin}^{2} \left(y\right)}}$

Finally, use equation $\textcolor{\mathmr{and} a n \ge}{\left(1\right)}$ to get

$\frac{\mathrm{df}}{d \theta} = \frac{9}{2} \cdot \frac{\sin {\left(9 \theta\right)}^{- \frac{1}{2}} \cdot \cos \left(9 \theta\right)}{\sqrt{1 - {\left(\sqrt{\sin \left(9 \theta\right)}\right)}^{2}}}$

This is equivalent to

$\frac{\mathrm{df}}{d \theta} = \textcolor{g r e e n}{\frac{9}{2} \cdot \cos \frac{9 \theta}{\sqrt{\sin \left(9 \theta\right)} \cdot \sqrt{1 - \sin \left(9 \theta\right)}}}$