Practice for Exam Two

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Practice for Exam Two.

1 : Consider the function: $f(x)=\sage {p1f1}$ . Which of the following is the correct setup to compute the derivative of $f(x)$ at $x=\sage {p1target}$ ?

1:: $\lim \limits _{h\rightarrow 0}\frac {\sage {p1ans1}}{h}$
2:: $\lim \limits _{h\rightarrow 0}\frac {\sage {p1ans2}}{h}$
3:: $\lim \limits _{h\rightarrow 0}\frac {\sage {p1ans3}}{h}$
4:: $\lim \limits _{h\rightarrow 0}\frac {\sage {p1ans4}}{h}$

$\answer {\sage {p1correct}}$ .

If we want to evaluate the derivative of $f(x)$ at $\sage {p1target}$ , then we want to solve: $\lim \limits _{h\rightarrow 0} \frac {f(\sage {p1target}+h) - f(\sage {p1target})}{h}$ .

2 : Compute the formula for the line that is tangent to the function $f(x) = \sage {p2f1}$ at $x=\sage {p2target}$ . $y = \answer {\sage {p2ans}}$

First find the derivative of $f(x)$ at $\sage {p2target}$ and use that as the slope in your line formula; the easiest line formula to use is probably $y=m(x-x_1)+y_1$ for a point $(x_1,y_1)$ .

3 : Consider $f(x) = \sage {p3f1}$ . Compute the derivative of $f(x)$ using the limit definition of $f(x)$ . $f'(x) = \answer {\sage {p3ans}}$

You can always use your derivative rules to double check your answer, but keep in mind that actual exams will likely have variations that may not allow that checking so it is a good idea to do this problem as the limit form. In particular, you want to solve: $f'(x) = \lim \limits _{\Delta x\rightarrow 0} \frac {f(x + \Delta x) - f(x)}{\Delta x}$

4 : Let $f(x) = \sage {p4f1}$ . Compute the derivative of $f(x)$ using polynomial rules. $f'(x)=\answer {\sage {p4ans}}$

You want to solve this derivative using the polynomial rule; $\frac {d}{dx}\left [x^n\right ] = nx^{n-1}$ .

5 : Let $f(x) = \sage {p5f1}$ . Compute the derivative of $f(x)$ using polynomial rules. $f'(x)=\answer {\sage {p5ans}}$

Remember that the polynomial rule works for any power, not just whole numbers... so you can use it to solve radicals as well. Thus you want to solve this derivative using the polynomial rule; $\frac {d}{dx}\left [x^n\right ] = nx^{n-1}$ .

6 : Let $f(x) = \sage {p6f1}$ . Compute the derivative of $f(x)$ using exponential rules. $f'(x)=\answer {\sage {p6ans}}$

Remember your exponential rule, $\frac {d}{dx} \left [a^x\right ] = \ln (a)a^x$

7 : Let $f(x) = \log _{\sage {p7base}}(x)$ . Compute the derivative of $f(x)$ using logarithm rules. $f'(x)=\answer {\sage {p7ans}}$

Remember your logarithm rule, $\frac {d}{dx} \left [log_a(x)\right ] = \frac {1}{x\ln (a)}$

8 : Compute the derivative of $f(x) = \sage {p8fDisp}$ $f'(x) = \answer {\sage {p8ans}}$

Don’t forget you need to use the product rule when you have the product of two functions!

9 : Compute the derivative of $f(x) = \sage {p9fDisp}$ $f'(x) = \answer {\sage {p9ans}}$

Don’t forget you need to use the quotient rule (or product and chain rules) when you have the quotient of two functions!

10 : Compute the derivative of $f(x) = \sage {p10fDisp}$ $f'(x) = \answer {\sage {p10ans}}$

Don’t forget you need to use the chain rule when you have the composition of two (or more) functions. Try thinking of the outer function as $\sage {p10f1}$ and the inner function as $\sage {p10f2}$ . (Note: Due to the randomized nature of this problem, there may be much simpler outer and inner function options, these are just two possibilities)

11 : Suppose that we have the following implicitly defined function, where $y$ is a function of $x$ : $\sage {p11fL} = \sage {p11fR}$

Compute $\frac {dy}{dx}$ explicitly using implicit differentiation. $\frac {dy}{dx} = \answer {\sage {p11ans}}$

Remember you want to take the derivative using implicit differentiation, then solve for the $\frac {dy}{dx}$ terms that appear as a consequence of the chain rule applied to $y$ .

12 : Compute the derivative of $f(x) = \sage {p12fDisp}$ . $f'(x) = \answer {\sage {p12ans}}$

When faced with a large number of factors all being multiplied, using logarithmic differentiation is a good idea!