find the derivative of the function. f(θ) = arcsin(√sin(7θ)) f(θ) =

Explanation:

Step1: Let $u = \sqrt{\sin(7\theta)}$

$F(\theta)=\arcsin(u)$

Step2: Find derivative of outer - function

The derivative of $y = \arcsin(u)$ with respect to $u$ is $\frac{1}{\sqrt{1 - u^{2}}}$. So, $\frac{dF}{du}=\frac{1}{\sqrt{1 - u^{2}}}$.

Step3: Find derivative of inner - function $u=\sqrt{\sin(7\theta)}=(\sin(7\theta))^{\frac{1}{2}}$

Let $v=\sin(7\theta)$. Then $u = v^{\frac{1}{2}}$. First, $\frac{du}{dv}=\frac{1}{2}v^{-\frac{1}{2}}=\frac{1}{2\sqrt{v}}$. Second, $\frac{dv}{d\theta}=7\cos(7\theta)$. By the chain - rule $\frac{du}{d\theta}=\frac{du}{dv}\cdot\frac{dv}{d\theta}=\frac{7\cos(7\theta)}{2\sqrt{\sin(7\theta)}}$.

Step4: Use the chain - rule $\frac{dF}{d\theta}=\frac{dF}{du}\cdot\frac{du}{d\theta}$

Substitute $u = \sqrt{\sin(7\theta)}$ into $\frac{dF}{du}$ and $\frac{du}{d\theta}$ we found above. $\frac{dF}{d\theta}=\frac{1}{\sqrt{1 - (\sqrt{\sin(7\theta)})^{2}}}\cdot\frac{7\cos(7\theta)}{2\sqrt{\sin(7\theta)}}=\frac{7\cos(7\theta)}{2\sqrt{\sin(7\theta)}\sqrt{1 - \sin(7\theta)}}$

Answer:

$\frac{7\cos(7\theta)}{2\sqrt{\sin(7\theta)}\sqrt{1 - \sin(7\theta)}}$