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AP Calculus BC : Derivative at a Point

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Example Questions

Example Question #21 : Derivative At A Point

Calculate the derivative of $f(x)=\frac{x^3}{3}-2x^2+x-3$ $\displaystyle f(x)=\frac{x^3}{3}-2x^2+x-3$ at the point x=3 $\displaystyle x=3$.

Possible Answers:

$\displaystyle -3$

$\displaystyle -4$

$\displaystyle -2$

$\displaystyle -1$

Correct answer:

$\displaystyle -2$

Explanation:

There are 2 steps to solving this problem.

First, take the derivative of f(x) $\displaystyle f(x)$

Then, replace the value of x with the given point and evaluate

For example, if x=a $\displaystyle x=a$ , then we are looking for the value of f'(x=a) $\displaystyle f'(x=a)$ , or the derivative of f(x) $\displaystyle f(x)$ at x=a $\displaystyle x=a$.

$f(x)=\frac{x^3}{3}-2x^2+x-3$ $\displaystyle f(x)=\frac{x^3}{3}-2x^2+x-3$

Calculate f'(x) $\displaystyle f'(x)$

Derivative rules that will be needed here:

Derivative of a constant is 0. For example, $\frac{\mathrm{d} }{\mathrm{d} t} 5 = 0$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} t} 5 = 0$
Taking a derivative on a term, or using the power rule, can be done by doing the following: $\frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$ $\displaystyle \frac{\mathrm{d} }{\mathrm{d} t} t^n = n * t^{n-1}$

$\displaystyle f'(x)=x^2 - 4x + 1$

Then, plug in the value of x and evaluate

$\displaystyle f'(x=3)=3^2-4*3+1=-2$

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Example Question #12 : Derivative Defined As Limit Of Difference Quotient

If $f(x)=\ln(2x)$ $\displaystyle f(x)=\ln(2x)$, which of the following limits equals f'(e+1) $\displaystyle f'(e+1)$?

Possible Answers:

$\lim_{h \to \infty} \frac{\ln(e+1+h)-\ln(e+1)}{e+1}$ $\displaystyle \lim_{h \to \infty} \frac{\ln(e+1+h)-\ln(e+1)}{e+1}$

$\lim_{h \to \infty} \frac{\ln(e+1+h)-\ln(e+1)}{h}$ $\displaystyle \lim_{h \to \infty} \frac{\ln(e+1+h)-\ln(e+1)}{h}$

$\lim_{h \to 0} \frac{\ln(1+h)-\ln(h)}{h}$ $\displaystyle \lim_{h \to 0} \frac{\ln(1+h)-\ln(h)}{h}$

$\lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{h}$ $\displaystyle \lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{h}$

$\lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{e+1}$ $\displaystyle \lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{e+1}$

Correct answer:

$\lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{h}$ $\displaystyle \lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{h}$

Explanation:

The equation for the derivative at a point is given by

$f'(a)=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$ $\displaystyle f'(a)=\lim_{h \to 0}\frac{f(a+h)-f(a)}{h}$.

By substituting a = e+1 $\displaystyle a = e+1$, $f = \ln(2x)$ $\displaystyle f = \ln(2x)$, we obtain

$f'(e+1)=\lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{h}.$ $\displaystyle f'(e+1)=\lim_{h \to 0} \frac{\ln(e+1+h)-\ln(e+1)}{h}.$

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AP Calculus BC : Derivative at a Point

Study concepts, example questions & explanations for AP Calculus BC

All AP Calculus BC Resources

Example Questions

Example Question #21 : Derivative At A Point

Example Question #12 : Derivative Defined As Limit Of Difference Quotient

All AP Calculus BC Resources

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