Question

find the derivative of the following function by first expanding or simplifying the expression. $g(w)=\frac{4e^{2w}+e^{w}}{e^{w}}$ $g(w)=square$

Explanation:

Step1: Simplify the function

Use the rule $\frac{a + b}{c}=\frac{a}{c}+\frac{b}{c}$. So $g(w)=\frac{4e^{2w}+e^{w}}{e^{w}}=\frac{4e^{2w}}{e^{w}}+\frac{e^{w}}{e^{w}}$. By the rule of exponents $\frac{e^{m}}{e^{n}}=e^{m - n}$, we get $g(w)=4e^{w}+1$.

Step2: Differentiate the simplified - function

The derivative of $y = ae^{x}$ with respect to $x$ is $y'=ae^{x}$ and the derivative of a constant $C$ with respect to $x$ is $0$. For $g(w)=4e^{w}+1$, using the sum - rule of differentiation $(u + v)'=u'+v'$, where $u = 4e^{w}$ and $v = 1$. The derivative of $4e^{w}$ is $4e^{w}$ and the derivative of $1$ is $0$. So $g'(w)=4e^{w}$.

Answer:

$4e^{w}$