Question

simplify the function before differentiating

$f(x)=\frac{e^{x}+6 e^{9 x}}{e^{x}}$

$f(x)=1+6 e^{8 x}$ (simplify your answer)

$\frac{d}{d x}\left(\frac{e^{x}+6 e^{9 x}}{e^{x}}\
ight)=\square$

Explanation:

Step1: Simplify the function

We have \(f(x)=\frac{e^{x}+6e^{9x}}{e^{x}}\). Using the rule \(\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}\), we get \(f(x)=\frac{e^{x}}{e^{x}}+\frac{6e^{9x}}{e^{x}}\).
Since \(\frac{e^{m}}{e^{n}}=e^{m - n}\), then \(\frac{e^{x}}{e^{x}}=e^{x-x}=e^{0}=1\) and \(\frac{6e^{9x}}{e^{x}}=6e^{9x - x}=6e^{8x}\). So \(f(x)=1 + 6e^{8x}\).

Step2: Differentiate the simplified function

The derivative of a constant \(C\) is \(0\), i.e., \(\frac{d}{dx}(C)=0\). For the function \(y = e^{ax}\), its derivative is \(y^\prime=ae^{ax}\) (by the chain - rule, if \(u = ax\), \(y = e^{u}\), then \(\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}\), \(\frac{dy}{du}=e^{u}\), \(\frac{du}{dx}=a\), so \(\frac{dy}{dx}=ae^{ax}\)).
For \(y = f(x)=1 + 6e^{8x}\), using the sum rule \(\frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx}\) where \(u = 1\) and \(v = 6e^{8x}\).
\(\frac{d}{dx}(1)=0\) and \(\frac{d}{dx}(6e^{8x})\). Let \(a = 8\) in the formula \(\frac{d}{dx}(e^{ax})\), then \(\frac{d}{dx}(6e^{8x})=6\times8e^{8x}\) (by the constant multiple rule \(\frac{d}{dx}(k\cdot g(x))=k\cdot\frac{d}{dx}(g(x))\) where \(k = 6\) and \(g(x)=e^{8x}\)).

Answer:

\(48e^{8x}\)