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

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

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

Answer

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})