Section 1.1 Practice Gateway Exam 1
The use of a calculator or any computer algebra system is not permitted on this test. No partial credit is available. A score of at least 7 correct is required to pass this gateway.
Find the derivative of each of the following functions:
Exercises Exercises
1.
\(f(x) = x^4 + 2x^2 + \dfrac{1}{\sqrt{x}}\)
Solution.
\begin{align*}
f'(x) \amp = \frac{d}{dx} [x^4+2x^2+x^{-1/2}]\\
\amp = 4x^3 + 4x - \frac{1}{2} x^{-3/2}
\end{align*}
2.
\(g(x) = \dfrac{3x-\cos(x)}{x+1}\)
Solution.
\begin{align*}
g'(x) \amp = \frac{(x+1)\cdot\frac{d}{dx}[3x-\cos(x)] - (3x-\cos(x)) \cdot \frac{d}{dx}[x+1]}{(x+1)^2}\\
\amp = \frac{(x+1)\cdot(3+\sin(x)) - (3x-\cos(x))}{(x+1)^2}
\end{align*}
3.
\(r(x) = \ln(x^4+2)\)
Solution.
\begin{align*}
r'(x) \amp = \frac{1}{x^4+2} \cdot \frac{d}{dx}[x^4+2]\\
\amp = \frac{4x^3}{x^4+2}
\end{align*}
4.
\(y = \arcsin(3t^2+1)\)
Solution.
\begin{align*}
\frac{dy}{dt} \amp = \frac{1}{\sqrt{1-(3t^2+1)^2}}\cdot \frac{d}{dt}[3t^2+1]\\
\amp = \frac{6t}{\sqrt{1-(3t^2+1)^2}}
\end{align*}
5.
\(f(t) = 5^{\sin(t)}\)
Solution.
\begin{align*}
f'(t) \amp = \ln(5)5^{\sin(t)}\cdot \frac{d}{dt}[\sin(t)]\\
\amp = \ln(5) 5^{\sin(t)}\cos(t)
\end{align*}
6.
\(g(x) = (x^4-4x)^{12}\)
Solution.
\begin{align*}
g'(x) \amp = 12(x^4-4x)^{11} \cdot \frac{d}{dx}[x^4-4x]\\
\amp = 12(x^4-4x)^{11} \cdot (4x^3-4)
\end{align*}
7.
\(h(t) = e^t \cos(2t)\)
Solution.
\begin{align*}
h'(t) \amp = e^t \cdot \frac{d}{dt}[\cos(2t)] + \cos(2t) \cdot \frac{d}{dt}[e^t]\\
\amp = (e^t)(-\sin(2t)\cdot 2) + \cos(2t)e^{t}
\end{align*}
8.
\(h(x) = \dfrac{1}{1+x^2}\)
Solution.
Solution 1: Rewrite and use the chain rule.
\begin{align*}
h'(x) \amp = \frac{d}{dx}[(1+x^2)^{-1}]\\
\amp = -(1+x^2)^{-2} \cdot \frac{d}{dx}[1+x^2]\\
\amp = -(1+x^2)^{-2}\cdot 2x
\end{align*}
Solution 2: Use the quotient rule.
\begin{align*}
h'(x) \amp = \frac{(1+x^2) \cdot \frac{d}{dx}[1] - 1 \cdot \frac{d}{dx}[1+x^2]}{(1+x^2)^2}\\
\amp = \frac{(1+x^2)\cdot 0 - 1 \cdot 2x}{(1+x^2)^2}\\
\amp = \frac{-2x}{(1+x^2)^2}
\end{align*}