Practice Problems 9

Lecture 9: Continuity and Rules of Differentiation

Problem numbers refer to the textbook (Chiang & Wainwright, 4th edition). These are not graded, but you should attempt them all before the quiz. Try each question yourself first, then expand the Solution to check.

Exercise 7.1

3. Find $f^{\prime }(1)$ and $f^{\prime }(2)$ for the following functions:

1. $y=f(x)=18x$

Solution

$$f^{\prime }(x)=18$$

$$f^{\prime }(1)=18f^{\prime }(2)=18$$

2. $y=f(x)=c{x}^3$

Solution

$$f^{\prime }(x)=3c{x}^2$$

$$f^{\prime }(1)=3c{f}^{\prime }(2)=12c$$

3. $f(x)=-5x^{-2}$

Solution

$$f^{\prime }(x)=10x^{-3}$$

$$f^{\prime }(1)=10f^{\prime }(2)=\frac{10}{8}=\frac{5}{4}$$

4. $f(x)=\frac{3}{4}{x}^{4/3}$

Solution

$$f^{\prime }(x)=x^{1/3}=\sqrt[3]{x}$$

$$f^{\prime }(1)=1f^{\prime }(2)=\sqrt[3]{2}$$

5. $f(w)=6w^{1/3}$

Solution

$$f^{\prime }(w)=2w^{-2/3}$$

$$f^{\prime }(1)=2f^{\prime }(2)=2\cdot 2^{-2/3}=2^{1/3}$$

6. $f(w)=-3w^{-1/6}$

Solution

$$f^{\prime }(w)=\frac{1}{2}{w}^{-7/6}$$

$$f^{\prime }(1)=\frac{1}{2}{f}^{\prime }(2)=\frac{1}{2}\cdot 2^{-7/6}=2^{-13/6}$$

Exercise 7.2

3. Differentiate the following by using the product rule:

4. $(ax-b)(c{x}^2)$

Solution

Let $f(x)=ax-b$ and $g(x)=c{x}^2$, so $f^{\prime }(x)=a$ and $g^{\prime }(x)=2cx$.

$$\begin{array}{rl}\frac{d}{dx}[f(x)g(x)]& =f^{\prime }(x)g(x)+f(x)g^{\prime }(x)\\ & =ac{x}^2+(ax-b)2cx\\ & =ac{x}^2+2ac{x}^2-2bcx\\ & =3ac{x}^2-2bcx\end{array}$$

5. $(2-3x)(1+x)(x+2)$

Solution

Let $f(x)=(2-3x)(1+x)$ and $g(x)=x+2$, so $g^{\prime }(x)=1$. Using the product rule on $f(x)$ itself:

$$f^{\prime }(x)=-3(1+x)+(2-3x)(1)=-(6x+1)$$

Then

$$\begin{array}{rl}\frac{d}{dx}[f(x)g(x)]& =f^{\prime }(x)g(x)+f(x)g^{\prime }(x)\\ & =-(6x+1)(x+2)+(2-3x)(1+x)\\ & =-6x^2-13x-2+2-x-3x^2\\ & =-9x^2-14x=-x(9x+14)\end{array}$$

7. Find the derivatives of:

1. $(x^2+3)/x$

Solution

With $f(x)=x^2+3$, $f^{\prime }(x)=2x$, and $g(x)=x$, $g^{\prime }(x)=1$:

$$\begin{array}{rl}\frac{d}{dx}\frac{f(x)}{g(x)}& =\frac{f^{\prime }(x)g(x)-f(x)g^{\prime }(x)}{g(x{)}^2}\\ & =\frac{(2x)x-(x^2+3)\cdot 1}{x^2}=\frac{x^2-3}{x^2}\end{array}$$

2. $(x+9)/x$

Solution

$$\frac{d}{dx}\frac{x+9}{x}=\frac{1\cdot x-1\cdot (x+9)}{x^2}=\frac{-9}{x^2}$$

3. $6x/(x+5)$

Solution

$$\frac{d}{dx}\frac{6x}{x+5}=\frac{6(x+5)-1\cdot 6x}{(x+5{)}^2}=\frac{30}{(x+5{)}^2}$$

4. $(a{x}^2+b)/(cx+d)$

Solution

$$\begin{array}{rl}\frac{d}{dx}\frac{a{x}^2+b}{cx+d}& =\frac{2ax(cx+d)-(a{x}^2+b)c}{(cx+d{)}^2}\\ & =\frac{2ac{x}^2+2adx-ac{x}^2-bc}{(cx+d{)}^2}\\ & =\frac{ac{x}^2+2adx-bc}{(cx+d{)}^2}\end{array}$$

8. Given the function $f(x)=ax+b$, find the derivatives of:

1. $f(x)$

Solution

$$f^{\prime }(x)=a$$

2. $xf(x)$

Solution

$$\frac{d}{dx}(a{x}^2+bx)=2ax+b$$

3. $1/f(x)$

Solution

$$\frac{d}{dx}\frac{1}{ax+b}=\frac{0\cdot (ax+b)-a\cdot 1}{(ax+b{)}^2}=\frac{-a}{(ax+b{)}^2}$$

4. $f(x)/x$

Solution

$$\frac{d}{dx}\frac{ax+b}{x}=\frac{ax-(ax+b)\cdot 1}{x^2}=\frac{-b}{x^2}$$

Exercise 7.3

1. Given $y=u^3+2u$, where $u=5-x^2$, find $dy/dx$ by the chain rule.

Solution

$$\frac{dy}{du}=3u^2+2\frac{du}{dx}=-2x$$

$$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}=(3u^2+2)(-2x)=-2x[3{(5-x^2)}^2+2]$$

2. Given $w=a{y}^2$ and $y=b{x}^2+cx$, find $dw/dx$ by the chain rule.

Solution

$$\begin{array}{rl}\frac{dw}{dx}& =\frac{dw}{dy}\cdot \frac{dy}{dx}\\ & =2ay(2bx+c)\\ & =2a(b{x}^2+cx)(2bx+c)\end{array}$$

3. Use the chain rule to find $dy/dx$ for the following:

1. $y=(3x^2-13{)}^3$

Solution

Let $z=3x^2-13$, so $y=z^3$:

$$\frac{dy}{dx}=\frac{dy}{dz}\cdot \frac{dz}{dx}=3z^2(6x)=18x{(3x^2-13)}^2$$

2. $y=(7x^3-5{)}^9$

Solution

Let $z=7x^3-5$, so $y=z^9$:

$$\frac{dy}{dx}=\frac{dy}{dz}\cdot \frac{dz}{dx}=9z^8\left(21x^2\right)=189x^2{(7x^3-5)}^8$$

3. $y=(ax+b{)}^5$

Solution

Let $z=ax+b$, so $y=z^5$:

$$\frac{dy}{dx}=\frac{dy}{dz}\cdot \frac{dz}{dx}=5z^4(a)=5a(ax+b{)}^4$$

4. Given $y=(16x+3{)}^{-2}$, use the chain rule to find $dy/dx$. Then rewrite the function as $y=1/(16x+3{)}^2$ and find $dy/dx$ by the quotient rule. Are the answers identical?

Solution

Chain rule, with $z=16x+3$ and $y=z^{-2}$:

$$\frac{dy}{dx}=\frac{dy}{dz}\cdot \frac{dz}{dx}=-2z^{-3}(16)=\frac{-32}{(16x+3{)}^3}$$

Quotient rule, where the derivative of the denominator is $2(16x+3)\cdot 16=32(16x+3)$:

$$\frac{dy}{dx}=\frac{0\cdot (16x+3{)}^2-1\cdot 32(16x+3)}{(16x+3{)}^4}=\frac{-32}{(16x+3{)}^3}$$

The answers are identical.

5. Given $y=7x+21$, find its inverse function. Then find $dy/dx$ and $dx/dy$, and verify the inverse-function rule. Also verify that the graphs of the two functions bear a mirror-image relationship to each other.

Solution

The inverse function is $x={\displaystyle \frac{y-21}{7}}$.

$$\frac{dy}{dx}=7\frac{dx}{dy}=\frac{1}{7}=\frac{1}{dy/dx}$$

so the inverse-function rule holds. To graph the inverse with its own independent variable on the horizontal axis, write it as $y=(x-21)/7$. The two lines are mirror images of each other across the 45-degree line $y=x$:

6. Are the following functions strictly monotonic? For each strictly monotonic function, find $dx/dy$ by the inverse-function rule.

1. $y=-x^6+5(x>0)$

Solution

$$\frac{dy}{dx}=-6x^5<0\text{ for }x>0$$

So the function is strictly decreasing, hence strictly monotonic, and

$$\frac{dx}{dy}=\frac{1}{dy/dx}=\frac{-1}{6x^5}$$

2. $y=4x^5+x^3+3x$

Solution

$$\frac{dy}{dx}=20x^4+3x^2+3>0\text{ for all }x$$

So the function is strictly increasing, hence strictly monotonic, and

$$\frac{dx}{dy}=\frac{1}{20x^4+3x^2+3}$$