Question

find $\frac{ds}{dt}$. s = 3tan t - $e^{-t}$ $\frac{ds}{dt}=square$

Explanation:

Step1: Differentiate term - by - term

Use the derivative rules for $\tan t$ and $e^{-t}$.

Step2: Derivative of $3\tan t$

The derivative of $\tan t$ is $\sec^{2}t$, so the derivative of $3\tan t$ is $3\sec^{2}t$ by the constant - multiple rule ($(cf(x))' = cf'(x)$ where $c = 3$ and $f(t)=\tan t$).

Step3: Derivative of $-e^{-t}$

The derivative of $e^{u}$ with respect to $t$ is $e^{u}\frac{du}{dt}$ by the chain - rule. Here $u=-t$, so $\frac{du}{dt}=- 1$. Then the derivative of $-e^{-t}$ is $-e^{-t}\times(-1)=e^{-t}$.

Step4: Combine the derivatives

$\frac{ds}{dt}$ of $s = 3\tan t-e^{-t}$ is the sum of the derivatives of its terms. So $\frac{ds}{dt}=3\sec^{2}t + e^{-t}$.

Answer:

$3\sec^{2}t + e^{-t}$