Question

given ( y = f(u) ) and ( u = g(x) ), find ( \frac{dy}{dx}=f(g(x))g(x) ) for the following functions.
( y=sin u, u = 4x - 3 )
( \frac{dy}{dx}=f(g(x))g(x)=square )

Explanation:

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

Given \(y=\sin u\), by the derivative formula \((\sin x)^\prime=\cos x\), we have \(f^\prime(u)=\cos u\). Then \(f^\prime(g(x))=\cos(4x - 3)\) (since \(u = g(x)=4x-3\)).

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

Given \(u = 4x-3\), by the power rule \((ax + b)^\prime=a\) (\(a = 4\), \(b=-3\)), we have \(g^\prime(x)=4\).

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

Substitute \(f^\prime(g(x))=\cos(4x - 3)\) and \(g^\prime(x)=4\) into the formula \(\frac{dy}{dx}=f^\prime(g(x))g^\prime(x)\), we get \(\frac{dy}{dx}=4\cos(4x - 3)\).

Answer:

\(4\cos(4x - 3)\)