Chapter 3.3, Problem 24E

(a)

To determine

To calculate: The value of $\frac{d}{dx}\left(uv\right)$ at $x=2$ .

Answer to Problem 24E

The required value of $\frac{d}{dx}\left(uv\right)$ is 2.

Explanation of Solution

Given information:

The given values:

  $u\left(2\right)=3$

  $u^{\prime }\left(2\right)=-4$

  $v\left(2\right)=1$

  $v^{\prime }\left(2\right)=2$

Formula used:

Product rule:

  $\frac{d}{dx}\left(uv\right)=u{v}^{\prime }+v{u}^{\prime }$

Calculation:

The given expression is $\frac{d}{dx}\left(uv\right)$ .

Apply the product rule of differentiation on the given expression.

  $\frac{d}{dx}\left(uv\right)=u{v}^{\prime }+v{u}^{\prime }$

At $x=2$ , substitute 3 for $u$ , $-2$ for $u^{\prime }$ , 1 for $v$ and $2$ for $v^{\prime }$ in the above expression.

  $\begin{array}{c}\frac{d}{dx}\left(uv\right)=\left(3\right)\left(2\right)+\left(1\right)\left(-4\right)\\ =6-4\\ =2\end{array}$

Hence, the required value of $\frac{d}{dx}\left(uv\right)$ is 2.

(b)

To determine

To calculate: The value of $\frac{d}{dx}\left(\frac{u}{v}\right)$ at $x=2$ .

Answer to Problem 24E

The required value of $\frac{d}{dx}\left(\frac{u}{v}\right)$ is $-10$ .

Explanation of Solution

Given information:

The given values:

  $u\left(2\right)=3$

  $u^{\prime }\left(2\right)=-4$

  $v\left(2\right)=1$

  $v^{\prime }\left(2\right)=2$

Formula used:

Quotient rule:

  $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$

Calculation:

The given expression is $\frac{d}{dx}\left(\frac{u}{v}\right)$ .

Apply the quotient rule of differentiation on the given expression.

  $\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{u^{\prime }v-u{v}^{\prime }}{v^2}$

At $x=2$ , substitute 3 for $u$ , $-2$ for $u^{\prime }$ , 1 for $v$ and $2$ for $v^{\prime }$ in the above expression.

  $\begin{array}{c}\frac{d}{dx}\left(\frac{u}{v}\right)=\frac{\left(1\right)\left(-4\right)-\left(3\right)\left(2\right)}{{\left(1\right)}^2}\\ =\frac{-4-6}{1}\\ =-10\end{array}$

Hence, the required value of $\frac{d}{dx}\left(\frac{u}{v}\right)$ is $-10$ .

(c)

To determine

To calculate: The value of $\frac{d}{dx}\left(\frac{v}{u}\right)$ at $x=2$ .

Answer to Problem 24E

The required value of $\frac{d}{dx}\left(\frac{v}{u}\right)$ is $\frac{10}{9}$ .

Explanation of Solution

Given information:

The given values:

  $u\left(0\right)=5$

  $u^{\prime }\left(0\right)=-3$

  $v\left(0\right)=-1$

  $v^{\prime }\left(0\right)=2$

Formula used:

Quotient rule:

  $\frac{d}{dx}\left(\frac{v}{u}\right)=\frac{v^{\prime }u-u^{\prime }v}{u^2}$

Calculation:

The given expression is $\frac{d}{dx}\left(\frac{v}{u}\right)$ .

Apply the quotient rule of differentiation on the given expression.

  $\frac{d}{dx}\left(\frac{v}{u}\right)=\frac{v^{\prime }u-u^{\prime }v}{u^2}$

At $x=2$ , substitute 3 for $u$ , $-2$ for $u^{\prime }$ , 1 for $v$ and $2$ for $v^{\prime }$ in the above expression.

  $\begin{array}{c}\frac{d}{dx}\left(\frac{v}{u}\right)=\frac{\left(3\right)\left(2\right)-\left(1\right)\left(-4\right)}{3^2}\\ =\frac{6+4}{9}\\ =\frac{10}{9}\end{array}$

Hence, the required value of $\frac{d}{dx}\left(\frac{v}{u}\right)$ is $\frac{10}{9}$ .

(d)

To determine

To calculate: The value of $\frac{d}{dx}\left(3u-2u+2uv\right)$ at $x=2$ .

Answer to Problem 24E

The required value of $\frac{d}{dx}\left(3u-2u+2uv\right)$ is $20$ .

Explanation of Solution

Given information:

The given values:

  $u\left(0\right)=5$

  $u^{\prime }\left(0\right)=-3$

  $v\left(0\right)=-1$

  $v^{\prime }\left(0\right)=2$

Calculation:

The given expression is $\frac{d}{dx}\left(3u-2u+2uv\right)$ .

Differentiate the given expression.

  $\frac{d}{dx}\left(3u-2u+2uv\right)=3u^{\prime }-2u^{\prime }+2\left(u{v}^{\prime }+v{u}^{\prime }\right)$

At $x=2$ , substitute 3 for $u$ , $-2$ for $u^{\prime }$ , 1 for $v$ and $2$ for $v^{\prime }$ in the above expression.

  $\begin{array}{c}\frac{d}{dx}\left(3u-2u+2uv\right)=3\left(-4\right)-2\left(2\right)+2\left(\left(3\right)\left(2\right)+\left(1\right)\left(-4\right)\right)\\ =-12-4+2\left(6-4\right)\\ =-12-4+4\\ =-12\end{array}$

Hence, the required value of $\frac{d}{dx}\left(3u-2u+2uv\right)$ is $-12$ .