homework5: problem 38 (1 point) let $f(x)=sqrt{7x - sin^{2}(2x)}$. find $f(x)$. $f(x)=$

Explanation:

Step1: Identify the outer - function and inner - function

Let $u = 7x-\sin^{2}(2x)$, then $y = \sqrt{u}=u^{\frac{1}{2}}$.

Step2: Differentiate the outer - function with respect to $u$

Using the power rule $\frac{d}{du}(u^{n})=nu^{n - 1}$, we have $\frac{dy}{du}=\frac{1}{2}u^{-\frac{1}{2}}=\frac{1}{2\sqrt{u}}$.

Step3: Differentiate the inner - function with respect to $x$

First, $\frac{d}{dx}(7x)=7$. For $\frac{d}{dx}(\sin^{2}(2x))$, let $t=\sin(2x)$. Then $\sin^{2}(2x)=t^{2}$. By the chain - rule, $\frac{d}{dt}(t^{2}) = 2t$ and $\frac{d}{dx}(\sin(2x))=\cos(2x)\cdot2 = 2\cos(2x)$. So $\frac{d}{dx}(\sin^{2}(2x))=2\sin(2x)\cdot2\cos(2x)=4\sin(2x)\cos(2x)$. Then $\frac{du}{dx}=7 - 4\sin(2x)\cos(2x)$.

Step4: Use the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$

Substitute $u = 7x-\sin^{2}(2x)$ and the values of $\frac{dy}{du}$ and $\frac{du}{dx}$: $\frac{dy}{dx}=\frac{7 - 4\sin(2x)\cos(2x)}{2\sqrt{7x-\sin^{2}(2x)}}$.

Answer:

$\frac{7 - 4\sin(2x)\cos(2x)}{2\sqrt{7x-\sin^{2}(2x)}}$