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5. using the power rule and other basic derivative rules, find the foll…

Question

  1. using the power rule and other basic derivative rules, find the following: a. $\frac{d}{dx}(5x^{3}+\frac{2}{x^{3}}+picdot e^{2})$ b. $\frac{d}{dt}(7t^{3/2}-\frac{6}{t^{3/2}})$ c. $\frac{d}{dx}(\frac{4sqrt3{x^{2}}+2x^{2}+7}{x^{2}})$ (note: you shouldnt need to use the quotient rule here!)

Explanation:

Step1: Recall the power - rule for derivatives

The power - rule states that if $y = ax^n$, then $\frac{dy}{dx}=nax^{n - 1}$, and the derivative of a constant $C$ is $0$. Also, $\frac{d}{dx}(u + v+w)=\frac{du}{dx}+\frac{dv}{dx}+\frac{dw}{dx}$.

Step2: Solve part A

We have $y = 5x^{3}+\frac{2}{x^{3}}+\pi e^{2}=5x^{3}+2x^{- 3}+\pi e^{2}$.
$\frac{d}{dx}(5x^{3}+2x^{-3}+\pi e^{2})=\frac{d}{dx}(5x^{3})+\frac{d}{dx}(2x^{-3})+\frac{d}{dx}(\pi e^{2})$.
Using the power - rule: $\frac{d}{dx}(5x^{3}) = 3\times5x^{3 - 1}=15x^{2}$, $\frac{d}{dx}(2x^{-3})=-3\times2x^{-3 - 1}=-6x^{-4}$, and $\frac{d}{dx}(\pi e^{2}) = 0$.
So, $\frac{d}{dx}(5x^{3}+\frac{2}{x^{3}}+\pi e^{2})=15x^{2}-\frac{6}{x^{4}}$.

Step3: Solve part B

We have $y = 7t^{3/2}-\frac{6}{t^{3/2}}=7t^{3/2}-6t^{-3/2}$.
$\frac{d}{dt}(7t^{3/2}-6t^{-3/2})=\frac{d}{dt}(7t^{3/2})-\frac{d}{dt}(6t^{-3/2})$.
Using the power - rule: $\frac{d}{dt}(7t^{3/2})=\frac{3}{2}\times7t^{\frac{3}{2}-1}=\frac{21}{2}t^{\frac{1}{2}}$, $\frac{d}{dt}(6t^{-3/2})=-\frac{3}{2}\times6t^{-\frac{3}{2}-1}=-9t^{-\frac{5}{2}}$.
So, $\frac{d}{dt}(7t^{3/2}-\frac{6}{t^{3/2}})=\frac{21}{2}\sqrt{t}-\frac{9}{t^{5/2}}$.

Step4: Solve part C

First, rewrite $y=\frac{4\sqrt[3]{x^{2}}+2x^{2}+7}{x^{2}}=\frac{4x^{2/3}+2x^{2}+7}{x^{2}}=4x^{2/3 - 2}+2x^{2 - 2}+7x^{-2}=4x^{-4/3}+2 + 7x^{-2}$.
$\frac{d}{dx}(4x^{-4/3}+2 + 7x^{-2})=\frac{d}{dx}(4x^{-4/3})+\frac{d}{dx}(2)+\frac{d}{dx}(7x^{-2})$.
Using the power - rule: $\frac{d}{dx}(4x^{-4/3})=-\frac{4}{3}\times4x^{-\frac{4}{3}-1}=-\frac{16}{3}x^{-\frac{7}{3}}$, $\frac{d}{dx}(2) = 0$, $\frac{d}{dx}(7x^{-2})=-2\times7x^{-2 - 1}=-14x^{-3}$.
So, $\frac{d}{dx}(\frac{4\sqrt[3]{x^{2}}+2x^{2}+7}{x^{2}})=-\frac{16}{3x^{7/3}}-\frac{14}{x^{3}}$.

Answer:

A. $15x^{2}-\frac{6}{x^{4}}$
B. $\frac{21}{2}\sqrt{t}-\frac{9}{t^{5/2}}$
C. $-\frac{16}{3x^{7/3}}-\frac{14}{x^{3}}$