3.1: Derivative Rules

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Throughout this module, if something does not exist, write DNE in the answer box.

Recap Video

Take a look at the following video which recaps the ideas from the section.

Example Video

Below is a video showing a worked example.

Problems

Here are the derivative rules you should know from the section:

Let $f(x)$ and $g(x)$ be differentiable functions, and $c$ a constant.

(Addition) $[f(x)\pm g(x)]' = f'(x) \pm g'(x)$ .
(Scaling) $[cf(x)]' = c\,f'(x)$
(Power Rule) If $f(x) = x^n$ , then $f'(x) = nx^{n-1}$ .
(Exponentials) If $f(x) = e^x$ , then $f'(x) =e^x$ .

If $f(x) = x^7$ , then find $f'(x)$ .

With the limit definition, this would be a tad annoying. Instead, we will use the power rule with $n=\answer {7}$ , which gives an answer of $f'(x) = \answer {7x^6}$ .

If $f(x) =2x^4+8x-10$ , find $f'(x)$ .

We will apply all our derivative rules to get this. Notice

$\begin{eqnarray*} f'(x) &=& (2x^4+8x-10)' \\ &=& (2x^4)'+(8x)' - (10)' \quad \text{by addition rule} \\ &=& 2(x^4)' + 8(x)' - (10)' \quad \text{by scaling rule} \\ &=& \answer{8x^3+8}. \end{eqnarray*}$

(For the last term, it might help to write $10 = 10x^0$ and use the power rule to differentiate it.)

Let $f(x) = 2x^4+8x-10$ as in the previous problem.

The graph of $f(x)$ has a horizontal tangent line at $x=\answer {-1}$ .
To find where the tangent line is horizontal, set $f'(x) = 0$ .
The equation of the tangent line to the graph of $f(x)$ at $x=1$ is equal to $y = \answer {16x-16}.$

If $\displaystyle ef(x) = e^x+3x^3+\frac {2}{x^3}$ , then $f'(x) = \answer {e^x+9x^2-\frac {6}{x^4}}.$

Use the exponential rule for the first term and the power rule for the last two terms, writing $2/x^3$ as $2x^{-3}$ to help you.

If $f(x) = x^2(x-1)-e^x$ , then:

$f'(x) = \answer {3x^2-2x-e^x}$ .
Expand $f(x)$ to get $f(x) = x^3-x^2-e^x$ .
$f''(x) = \answer {6x-2-e^x}$ .
This means differentiate $f'(x)$ .
$f'''(x) = \answer {6-e^x}$ .

If $f(x) = e^x+3x$ , then the tangent line to the graph of $f(x)$ is parallel to the line $y = 5x+5$ at $x = \answer {\ln (2)}$ .

Find $f'(x)$ and see where this is equal to the slope of the parallel line.

Suppose $f(x) = Ax^2+Bx+C$ has a horizontal tangent line at the point $(1,2)$ and also passes through $(0,5)$ . Then $A = \answer {3}, \quad B= \answer {-6}, \quad C= \answer {5}.$

Since the graph passes through $(0,5)$ , we know $f(0) = 5$ . Plugging this into the expression for $f(x)$ , we get $C = \answer {5}$ . Since the graph passes through $(1,2)$ , we know $f(1) =\answer {2}$ . Plugging this into the expression for $f(x)$ gives $A+B+C = 2.$ Since $C = \answer {5}$ , this gives $A+B = \answer {-3}.$ We need some other equation to help us solve for $A$ and $B$ . In this case, we also know the graph has a horizontal tangent at $x=1$ . This means $f'(1) = \answer {0}$ . Notice $f'(x) = \answer {2Ax+B},$ so $f'(1) = \answer {2A+B}$ . Therefore, $2A+B = 0.$ We have a system of equations: $A+B = -3,$ $2A+B = 0.$ Solving for $A$ and $B$ gives $A = \answer {3}$ and $B = \answer {-6}$ .