Answer:
Problem 20)

Problem 21)
A)
The velocity function is:

The acceleration function is:

B)

Explanation:
Problem 20)
We want to differentiate the equation:

We can take the natural log of both sides. This yields:

Since ln(aᵇ) = bln(a):

Take the derivative of both sides with respect to x:
![\displaystyle (d)/(dx)\left[\ln y \right]=(d)/(dx)\left[x \ln \cos x\right]](https://img.qammunity.org/2022/formulas/mathematics/college/wr8gfvy8ondv0zqjidouaivhjoxpi415mm.png)
Implicitly differentiate the left and use the product rule on the right. Therefore:

Simplify:

Simplify and multiply both sides by y:

Since y = (cos x)ˣ:
Problem 21)
We are given the position function of a particle:

A)
Recall that the velocity function is the derivative of the position function. Hence:
![\displaystyle v(t)=s'(t)=(d)/(dt)[\sin(2\pi t)+2\cos(\pi t)]](https://img.qammunity.org/2022/formulas/mathematics/college/20nqboxa3q67bevswmhpw65b69gjxijz5a.png)
Differentiate:

The acceleration function is the derivative of the velocity function. Hence:
![\displaystyle a(t)=v'(t)=(d)/(dt)[2\pi(\cos(2\pi t)-\sin(\pi t))]](https://img.qammunity.org/2022/formulas/mathematics/college/c3ch4y8j5j307zyr5fr8b5a2vo17b50nz3.png)
Differentiate:
![\displaystyle \begin{aligned} a(t)&=2\pi[-2\pi\sin(2\pi t)-\pi\cos(\pi t)]\\&=-2\pi^2(2\sin(2\pi t)+\cos(\pi t))\end{aligned}](https://img.qammunity.org/2022/formulas/mathematics/college/hndwgh22n15lmmbumrsfppt0snfih5686f.png)
B)
The position at t = 0 will be:

The velocity at t = 0 will be:

And the acceleration at t = 0 will be:
