Question

given y = f(u) and u = g(x), find $\frac{dy}{dx}=f(g(x))g(x)$. y = 8u - 1, u = $(\frac{1}{4})x^{7}$ $\frac{dy}{dx}=square$

Explanation:

Step1: Find the derivative of y with respect to u

Given $y = 8u - 1$, then $y^\prime=f^\prime(u)=\frac{d}{du}(8u - 1)=8$.

Step2: Find the derivative of u with respect to x

Given $u=\frac{1}{4}x^{7}$, then $u^\prime = g^\prime(x)=\frac{d}{dx}(\frac{1}{4}x^{7})=\frac{1}{4}\times7x^{6}=\frac{7}{4}x^{6}$.

Step3: Apply the chain - rule

By the chain - rule $\frac{dy}{dx}=f^\prime(g(x))g^\prime(x)$. Substitute $f^\prime(u) = 8$ and $g^\prime(x)=\frac{7}{4}x^{6}$ into the chain - rule formula. Since $f^\prime(g(x)) = 8$ (because $f^\prime(u)$ is a constant), then $\frac{dy}{dx}=8\times\frac{7}{4}x^{6}$.

Answer:

$14x^{6}$