differentiate the following function. f(x)=3 + 5e^3x f(x)=□

Explanation:

Step1: Recall sum - rule of differentiation

The derivative of a sum of functions $u(x)+v(x)$ is $u'(x)+v'(x)$. Let $u(x) = 3$ and $v(x)=5e^{3x}$. So $f'(x)=u'(x)+v'(x)$.

Step2: Differentiate the constant function

The derivative of a constant $C$ is 0. Since $u(x) = 3$ (a constant), $u'(x)=0$.

Step3: Differentiate the exponential - function using chain - rule

The chain - rule states that if $y = f(g(x))$, then $y'=f'(g(x))\cdot g'(x)$. For $v(x)=5e^{3x}$, let $g(x)=3x$ and $f(u) = 5e^{u}$. Then $f'(u)=5e^{u}$ and $g'(x)=3$. So $v'(x)=5e^{3x}\cdot3 = 15e^{3x}$.

Step4: Combine the results

$f'(x)=u'(x)+v'(x)=0 + 15e^{3x}=15e^{3x}$.

Answer:

$15e^{3x}$