Question

calculate the derivative of the following function. f(t) = \frac{e^{9t}}{3 + e^{9t}} f(t) =

Explanation:

Step1: Identify quotient - rule

The quotient - rule states that if $y=\frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = e^{9t}$, $v=3 + e^{9t}$.

Step2: Find $u^\prime$

Using the chain - rule, if $u = e^{9t}$, let $u = e^{u_1}$ and $u_1 = 9t$. Then $\frac{du}{du_1}=e^{u_1}$ and $\frac{du_1}{dt}=9$. So $u^\prime=\frac{du}{dt}=\frac{du}{du_1}\cdot\frac{du_1}{dt}=9e^{9t}$.

Step3: Find $v^\prime$

Since $v = 3+e^{9t}$, and the derivative of a constant is 0, and using the chain - rule for $e^{9t}$ as above, $v^\prime = 9e^{9t}$.

Step4: Apply quotient - rule

$f^\prime(t)=\frac{u^\prime v - uv^\prime}{v^{2}}=\frac{9e^{9t}(3 + e^{9t})-e^{9t}\cdot9e^{9t}}{(3 + e^{9t})^{2}}$.

Step5: Simplify the expression

Expand the numerator: $9e^{9t}(3 + e^{9t})-9e^{18t}=27e^{9t}+9e^{18t}-9e^{18t}$.
So $f^\prime(t)=\frac{27e^{9t}}{(3 + e^{9t})^{2}}$.

Answer:

$\frac{27e^{9t}}{(3 + e^{9t})^{2}}$