## 170- Quiz 5

Quiz 5 is here.

Solutions follow.

Problem 1 asks to compute the derivative of $\sin^2(\sqrt x)$.

Let $f(x)=\sin^2(\sqrt x)=(\sin(\sqrt x))^2$. Note that we can write $f$ as a composition of functions, $f=g(h(j(x)))$, where $g(x)=x^2$, $h(x)=\sin(x)$, and $j(x)=\sqrt x$.

To compute $f'(x)$, we use the chain rule: $f'(x)=g'(h(j(x)))\cdot h'(j(x))\cdot j'(x).$

For this, we first need the derivatives of $g$, $h$ and $j$:

• $g'(x)=2x$, so $g'(h(j(x)))=2\sin(\sqrt x)$.
• $h'(x)=\cos(x)$, so $h'(j(x))=\cos(\sqrt x)$.
• $\displaystyle j'(x)=\frac12 x^{-1/2}$, since $\sqrt x= x^{1/2}$.

Combining these three results, we get: $\displaystyle f'(x)=2\sin(\sqrt x)\cdot\cos(\sqrt x)\cdot\frac 12 x^{-1/2}=\frac{\sin(2\sqrt x)}{2\sqrt x}$.

Problem 2 asks for the equation of the line tangent to $\cos^2 x-\sin^2(2x)$ at $x = \pi/2$.

For this, we first find a point on the line, and its slope. Since we are told that the line is tangent to the given curve when $x=\pi/2$, this means that the line goes through the point $(\pi/2, y)$, where the value of $y$ is found by replacing $x=\pi/2$ in the given function. This gives us $y=\cos^2(\pi/2)-\sin^2(\pi)=0-0=0.$

To find the slope of the tangent line, we recall that this is just the value of the derivative of the given function, when $x=\pi/2$. The derivative is given by $2\cos(x)\cdot(-\sin(x))-2\sin(2x)\cdot\cos(2x)\cdot 2.$

Again, this is $0$ when $x=\pi/2$.

This means that the tangent line we require goes through the point $(\pi/2,0)$ and has slope $0$, i.e., it is simply the line $y=0$.

Here is the graph of the function $\cos^2x-\sin^2(2x)$, with the tangent line and the line $x=\pi/2$ indicated. Click the image above to enlarge.