3.2 differentiate the following with respect to $x$: 3.2.1 $y = sqrt3{3x + 7}$. (2) 3.2.2 $y=ln(x^{2}cdot…

Explanation:

Step1: Rewrite the function for 3.2.1

Rewrite $y = \sqrt[3]{3x + 7}=(3x + 7)^{\frac{1}{3}}$. Using the chain - rule $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$, let $u = 3x+7$, then $y = u^{\frac{1}{3}}$. $\frac{dy}{du}=\frac{1}{3}u^{-\frac{2}{3}}$ and $\frac{du}{dx}=3$.

Step2: Calculate the derivative for 3.2.1

$\frac{dy}{dx}=\frac{1}{3}(3x + 7)^{-\frac{2}{3}}\cdot3=(3x + 7)^{-\frac{2}{3}}=\frac{1}{(3x + 7)^{\frac{2}{3}}}$

Step3: Simplify the function for 3.2.2

Use the property of logarithms $\ln(ab)=\ln a+\ln b$. So $y=\ln(x^{2}\cdot e^{x})=\ln x^{2}+\ln e^{x}=2\ln x + x$.

Step4: Calculate the derivative for 3.2.2

The derivative of $2\ln x$ is $\frac{2}{x}$ and the derivative of $x$ is 1. So $\frac{dy}{dx}=\frac{2}{x}+1=\frac{2 + x}{x}$

Step5: Use the quotient - rule for 3.2.3

The quotient - rule states that if $y=\frac{u}{v}$, then $\frac{dy}{dx}=\frac{u'v - uv'}{v^{2}}$. Here, $u=x - 1$, $u' = 1$, $v=x + 1$, $v'=1$.

Step6: Calculate the derivative for 3.2.3

$\frac{dy}{dx}=\frac{1\cdot(x + 1)-(x - 1)\cdot1}{(x + 1)^{2}}=\frac{x + 1-x + 1}{(x + 1)^{2}}=\frac{2}{(x + 1)^{2}}$

Answer:

3.2.1: $\frac{1}{(3x + 7)^{\frac{2}{3}}}$ 3.2.2: $\frac{x + 2}{x}$ 3.2.3: $\frac{2}{(x + 1)^{2}}$