Question

an $y = f(u)$ and $u = g(x)$, find $\frac{dy}{dx}=f(g(x))g(x)$.
$y = 8u^{8}$, $u = 4x + 19$

Explanation:

Step1: Find the derivative of \(y = f(u)\)

Given \(y = 8u^{8}\), using the power rule \(\frac{d}{du}(au^{n})=anu^{n - 1}\), we have \(f^{\prime}(u)=\frac{dy}{du}=8\times8u^{7}=64u^{7}\).

Step2: Find the derivative of \(u = g(x)\)

Given \(u = 4x + 19\), using the sum rule \(\frac{d}{dx}(ax + b)=a\), we have \(g^{\prime}(x)=\frac{du}{dx}=4\).

Step3: Apply the chain - rule \(\frac{dy}{dx}=f^{\prime}(g(x))g^{\prime}(x)\)

Substitute \(u = g(x)=4x + 19\) into \(f^{\prime}(u)\) and multiply by \(g^{\prime}(x)\).
\(f^{\prime}(g(x))=64(4x + 19)^{7}\), and \(g^{\prime}(x)=4\).
So \(\frac{dy}{dx}=64(4x + 19)^{7}\times4\).

Answer:

\(\frac{dy}{dx}=256(4x + 19)^{7}\)