Practice Problems 8

Lecture 8: Limits and the Derivative

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 6.2

2. Given the function $y=5x^2-4x$:

1. Find the difference quotient as a function of $x$ and $\mathrm{\Delta }x$.

Solution

$$\begin{array}{rl}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}& =\frac{f(x+\mathrm{\Delta }x)-f(x)}{\mathrm{\Delta }x}\\ & =\frac{5(x+\mathrm{\Delta }x{)}^2-4(x+\mathrm{\Delta }x)-(5x^2-4x)}{\mathrm{\Delta }x}\\ & =\frac{5(x^2+2x\mathrm{\Delta }x+(\mathrm{\Delta }x{)}^2)-4x-4\mathrm{\Delta }x-5x^2+4x}{\mathrm{\Delta }x}\\ & =\frac{10x\mathrm{\Delta }x+5(\mathrm{\Delta }x{)}^2-4\mathrm{\Delta }x}{\mathrm{\Delta }x}=10x-4+5\mathrm{\Delta }x\end{array}$$

2. Find the derivative $dy/dx$, using the limit definition.

Solution

$$\frac{dy}{dx}=\underset{\mathrm{\Delta }x\to 0}{lim}\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{lim}(10x-4+5\mathrm{\Delta }x)=10x-4$$

3. Find $f^{\prime }(2)$ and $f^{\prime }(3)$.

Solution

$$f^{\prime }(2)=10\times 2-4=16f^{\prime }(3)=10\times 3-4=26$$

3. Given the function $y=5x-2$:

1. Find the difference quotient $\mathrm{\Delta }y/\mathrm{\Delta }x$. What type of function is it?

Solution

$$\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\frac{5(x+\mathrm{\Delta }x)-2-(5x-2)}{\mathrm{\Delta }x}=\frac{5\mathrm{\Delta }x}{\mathrm{\Delta }x}=5$$

It is a constant function.

2. Since $\mathrm{\Delta }x$ does not appear in $\mathrm{\Delta }y/\mathrm{\Delta }x$ in part (a), does it make any difference to the value of $\mathrm{\Delta }y/\mathrm{\Delta }x$ whether $\mathrm{\Delta }x$ is large or small? Consequently, what is the limit of the difference quotient as $\mathrm{\Delta }x$ approaches zero?

Solution

No: the difference quotient equals 5 whatever the size of $\mathrm{\Delta }x$. So its limit as $\mathrm{\Delta }x\to 0$ is also 5, i.e. $dy/dx=5$.