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


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.


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