Question

question. given (y = f(u)) and (u = g(x)), find (\frac{dy}{dx}) by using leibnizs notation for the chain - rule: (\frac{dy}{dx}=\frac{dy}{du}cdot\frac{du}{dx}). (y = 4u^{4}+2), (u=sqrt3{x}). provide your answer below. (\frac{dy}{dx}=square)

Explanation:

Step1: Differentiate y with respect to u

Using the power - rule $\frac{d}{du}(au^n)=nau^{n - 1}$, for $y = 4u^4+2$, we have $\frac{dy}{du}=4\times4u^{4 - 1}+0 = 16u^3$.

Step2: Differentiate u with respect to x

Given $u=\sqrt[3]{x}=x^{\frac{1}{3}}$, by the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$, we get $\frac{du}{dx}=\frac{1}{3}x^{\frac{1}{3}-1}=\frac{1}{3}x^{-\frac{2}{3}}$.

Step3: Apply the chain - rule

The chain - rule is $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}$. Substitute $\frac{dy}{du}=16u^3$ and $\frac{du}{dx}=\frac{1}{3}x^{-\frac{2}{3}}$ into it. Since $u = x^{\frac{1}{3}}$, then $\frac{dy}{dx}=16(x^{\frac{1}{3}})^3\cdot\frac{1}{3}x^{-\frac{2}{3}}$.
Simplify the expression: $16x\cdot\frac{1}{3}x^{-\frac{2}{3}}=\frac{16}{3}x^{1-\frac{2}{3}}=\frac{16}{3}x^{\frac{1}{3}}$.

Answer:

$\frac{16}{3}\sqrt[3]{x}$