Question

differentiate the following function.
y = 8 ln x + ln 8
choose the correct setup below to start differentiating the function.
a. \\( \frac { d } { d x } ( 8 ln x + ln 8 ) = \frac { d } { d x } ( ln ( 8 x + 8 ) ) \\)
b. \\( \frac { d } { d x } ( 8 ln x + ln 8 ) = \frac { d } { d x } ( 8 ln x ) + \frac { d } { d x } ( ln 8 ) \\)
c. \\( \frac { d } { d x } ( 8 ln x + ln 8 ) = \frac { ln 8 cdot \frac { d } { d x } ( 8 ln x ) + 8 ln x cdot \frac { d } { d x } ( ln 8 ) } { ( ln 8 ) ^ { 2 } } \\)
d. \\( \frac { d } { d x } ( 8 ln x + ln 8 ) = ln 8 cdot \frac { d } { d x } ( 8 ln x ) + 8 ln x cdot \frac { d } { d x } ( ln 8 ) \\)
\\( \frac { d } { d x } ( 8 ln x + ln 8 ) = \\)

Explanation:

Step1: Apply sum rule of differentiation

The sum rule states that \(\frac{d}{dx}(u + v)=\frac{d}{dx}(u)+\frac{d}{dx}(v)\). For \(y = 8\ln x+\ln8\), let \(u = 8\ln x\) and \(v=\ln8\). So \(\frac{d}{dx}(8\ln x+\ln8)=\frac{d}{dx}(8\ln x)+\frac{d}{dx}(\ln8)\)

Step2: Differentiate \(8\ln x\)

Using the constant - multiple rule \(\frac{d}{dx}(k\cdot f(x))=k\cdot\frac{d}{dx}(f(x))\) and the formula \(\frac{d}{dx}(\ln x)=\frac{1}{x}\), we have \(\frac{d}{dx}(8\ln x)=8\cdot\frac{d}{dx}(\ln x)=8\cdot\frac{1}{x}=\frac{8}{x}\)

Step3: Differentiate \(\ln8\)

Since \(\ln8\) is a constant, and the derivative of a constant \(C\) is \(0\) (\(\frac{d}{dx}(C) = 0\)), so \(\frac{d}{dx}(\ln8)=0\)

Answer:

\(\frac{8}{x}\)