Chapter 4, Problem 60RE

a.

To determine

Find the derivative of each function easily by simplifying the expression before differentiating.

Answer to Problem 60RE

  $\frac{dy}{dx}=\frac{2}{2x+7}-\frac{3}{3x+2}.$

Explanation of Solution

Given information: Given function is $y=\ln (\frac{2x+7}{3x+2}).$

Calculation:

  $\begin{array}{l}y=\ln (\frac{2x+7}{3x+2}).\\ y=\ln (2x+7)-\ln (3x+2).\\ \frac{dy}{dx}=\frac{2}{2x+7}-\frac{3}{3x+2}.\end{array}$

b.

To determine

Find the derivative of each function easily by simplifying the expression before differentiating.

Answer to Problem 60RE

  $\frac{dy}{dx}=1.$

Explanation of Solution

Given information: Given function is $y=\frac{{(x^2-1)}^2}{(x^2-2x+1)(x+1)}.$

Calculation:

  $\begin{array}{l}y=\frac{{(x^2-1)}^2}{(x^2-2x+1)(x+1)}.\\ y=\frac{(x+1)(x+1)(x-1)(x-1)}{(x-1)(x-1)(x+1)}.\\ y=(x+1).\\ \frac{dy}{dx}=1.\end{array}$

c.

To determine

Find the derivative of each function easily by simplifying the expression before differentiating.

Answer to Problem 60RE

  $\frac{ds}{dt}=-2t.$

Explanation of Solution

Given information: Given function is $s={\sin }^2({\cos }^{-1}t).$

Calculation:

  $\begin{array}{l}s={\sin }^2({\cos }^{-1}t).\\ s=1-{\cos }^2({\cos }^{-1}t).\\ $ s=1-{[\cos ({\cos }^{-1}t)]}^2.$s=1-t^2.\\ \frac{ds}{dt}=0-2t.\\ \frac{ds}{dt}=-2t.\end{array}$

d.

To determine

Find the derivative of each function easily by simplifying the expression before differentiating.

Answer to Problem 60RE

  $\frac{ds}{dt}=\frac{80}{3\sqrt[6]{t}}.$

Explanation of Solution

Given information: Given function is $\begin{array}{}$ s={(\frac{2\sqrt{t}}{\sqrt[3]{t}})}^5.$\end{array}$

Calculation:

  $\begin{array}{}$ s={(\frac{2\sqrt{t}}{\sqrt[3]{t}})}^5.$$ s={(\frac{2t^{\frac{1}{2}}}{t^{\frac{1}{3}}})}^{5.}$$ s={(2t^{\frac{1}{2}-\frac{1}{3}})}^5.$$ s={(2t^{\frac{1}{6}})}^5=2^5.t^{\frac{5}{6}}=32.t^{\frac{5}{6}}.$$ \frac{ds}{dt}=32(\frac{5}{6})t^{-\frac{1}{6}}.$$ \frac{ds}{dt}=\frac{80}{3}{t}^{-\frac{1}{6}}.$$ \frac{ds}{dt}=\frac{80}{3\sqrt[6]{t}}.$\end{array}$