Question

find the derivative of y with respect to x.
y = ln(16x) + 4x
\frac{dy}{dx} = \square

Explanation:

Step1: Differentiate $\ln(16x)$

Use the chain - rule. If \(y = \ln(u)\) and \(u = 16x\), then \(\frac{dy}{du}=\frac{1}{u}\) and \(\frac{du}{dx}=16\). By the chain - rule \(\frac{d}{dx}(\ln(16x))=\frac{1}{16x}\times16=\frac{1}{x}\).

Step2: Differentiate \(4x\)

Use the power - rule \(\frac{d}{dx}(ax^{n})=anx^{n - 1}\). For \(y = 4x\) (where \(a = 4\) and \(n = 1\)), \(\frac{d}{dx}(4x)=4\times1\times x^{1-1}=4\).

Step3: Sum the derivatives

By the sum - rule of differentiation \(\frac{d}{dx}(f(x)+g(x))=\frac{d}{dx}(f(x))+\frac{d}{dx}(g(x))\). Here \(f(x)=\ln(16x)\) and \(g(x)=4x\), so \(\frac{dy}{dx}=\frac{1}{x}+4\).

Answer:

\(\frac{1}{x}+4\)