Practice Problems 10

Lecture 10: Exponential and Log Functions

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 10.5

1. Find the derivatives of:

1. $y=e^{2t+4}$

Solution

$$\frac{dy}{dt}=2e^{2t+4}$$

2. $y=e^{1-9t}$

Solution

$$\frac{dy}{dt}=-9e^{1-9t}$$

3. $y=e^{t^2+1}$

Solution

$$\frac{dy}{dt}=2t{e}^{t^2+1}$$

4. $y=5e^{2-t^2}$

Solution

$$\frac{dy}{dt}=-10t{e}^{2-t^2}$$

5. $y=e^{a{x}^2+bx+c}$

Solution

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

6. $y=x{e}^x$

Solution

$$\frac{dy}{dx}=x\frac{d}{dx}{e}^x+e^x\frac{d}{dx}x=x{e}^x+e^x=(x+1)e^x$$

7. $y=x^2{e}^{2x}$

Solution

$$\frac{dy}{dx}=x^2\left(2e^{2x}\right)+2x{e}^{2x}=2x(x+1)e^{2x}$$

8. $y=ax{e}^{bx+c}$

Solution

$$\frac{dy}{dx}=a(xb{e}^{bx+c}+e^{bx+c})=a(bx+1)e^{bx+c}$$

3. Find the derivatives of:

1. $y=\ln \left(7t^5\right)$

Solution

$$\frac{dy}{dt}=\frac{35t^4}{7t^5}=\frac{5}{t}$$

2. $y=\ln \left(a{t}^{10}\right)$

Solution

$$\frac{dy}{dt}=\frac{10a{t}^9}{a{t}^{10}}=\frac{10}{t}$$

3. $y=\ln (t+19)$

Solution

$$\frac{dy}{dt}=\frac{1}{t+19}$$

4. $y=5\ln (t+1{)}^2$

Solution

$$\frac{dy}{dt}=5\cdot \frac{2(t+1)}{(t+1{)}^2}=\frac{10}{t+1}$$

5. $y=\ln x-\ln (1+x)$

Solution

$$\frac{dy}{dx}=\frac{1}{x}-\frac{1}{1+x}=\frac{1}{x(1+x)}$$

6. $y=\ln [x(1-x{)}^8]$

Solution

$$\frac{dy}{dx}=\frac{d}{dx}[\ln x+8\ln (1-x)]=\frac{1}{x}-\frac{8}{1-x}=\frac{1-9x}{x(1-x)}$$

7. $y=\ln \left(\frac{2x}{1+x}\right)$

Solution

$$\frac{dy}{dx}=\frac{d}{dx}[\ln 2x-\ln (1+x)]=\frac{2}{2x}-\frac{1}{1+x}=\frac{1}{x(1+x)}$$

8. $y=5x^4\ln x^2$

Solution

$$\begin{array}{rl}\frac{dy}{dx}& =5x^4\cdot \frac{2x}{x^2}+20x^3\ln x^2\\ & =10x^3(1+2\ln x^2)=10x^3(1+4\ln x)\end{array}$$

7. Find the derivatives of the following by first taking the natural log of both sides:

1. $y={\displaystyle \frac{3x}{(x+2)(x+4)}}$

Solution

Since $\ln y=\ln 3x-\ln (x+2)-\ln (x+4)$,

$$\frac{1}{y}\frac{dy}{dx}=\frac{1}{x}-\frac{1}{x+2}-\frac{1}{x+4}=\frac{8-x^2}{x(x+2)(x+4)}$$

Hence,

$$\begin{array}{rl}\frac{dy}{dx}& =\frac{8-x^2}{x(x+2)(x+4)}\cdot \frac{3x}{(x+2)(x+4)}\\ & =\frac{3(8-x^2)}{(x+2{)}^2(x+4{)}^2}\end{array}$$

2. $y=(x^2+3)e^{x^2+1}$

Solution

Since $\ln y=\ln (x^2+3)+x^2+1$,

$$\frac{1}{y}\frac{dy}{dx}=\frac{2x}{x^2+3}+2x=\frac{2x(x^2+4)}{x^2+3}$$

Hence,

$$\frac{dy}{dx}=\frac{2x(x^2+4)}{x^2+3}\cdot (x^2+3)e^{x^2+1}=2x(x^2+4)e^{x^2+1}$$