Question

differentiate the following function. f(x)=9 - 2x + 3e^x f(x)=□

Explanation:

Step1: Differentiate constant term

The derivative of a constant $C$ is 0. For the constant 9 in $f(x)$, its derivative is 0 since $\frac{d}{dx}(9)=0$.

Step2: Differentiate linear - term

The derivative of $ax$ with respect to $x$ is $a$. For the term $- 2x$, using the power - rule $\frac{d}{dx}(x^n)=nx^{n - 1}$ (here $n = 1$), we have $\frac{d}{dx}(-2x)=-2$.

Step3: Differentiate exponential term

The derivative of $e^x$ with respect to $x$ is $e^x$. For the term $3e^x$, by the constant - multiple rule $\frac{d}{dx}(cf(x))=c\frac{d}{dx}(f(x))$ where $c = 3$ and $f(x)=e^x$, we get $\frac{d}{dx}(3e^x)=3\frac{d}{dx}(e^x)=3e^x$.

Step4: Combine the derivatives

Using the sum - rule of differentiation $\frac{d}{dx}(u + v+w)=\frac{d}{dx}(u)+\frac{d}{dx}(v)+\frac{d}{dx}(w)$, where $u = 9$, $v=-2x$, and $w = 3e^x$. So $f'(x)=\frac{d}{dx}(9)+\frac{d}{dx}(-2x)+\frac{d}{dx}(3e^x)=0-2 + 3e^x$.

Answer:

$-2 + 3e^x$