Question

find $\frac{ds}{dt}$. $s = t^{4}-3sec t + 5e^{t}$ $\frac{ds}{dt}=square$

Explanation:

Step1: Differentiate $t^4$

Using the power - rule $\frac{d}{dt}(t^n)=nt^{n - 1}$, for $n = 4$, we have $\frac{d}{dt}(t^4)=4t^{3}$.

Step2: Differentiate $-3\sec t$

The derivative of $\sec t$ is $\sec t\tan t$, so $\frac{d}{dt}(-3\sec t)=-3\sec t\tan t$.

Step3: Differentiate $5e^{t}$

The derivative of $e^{t}$ is $e^{t}$, so $\frac{d}{dt}(5e^{t}) = 5e^{t}$.

Step4: Combine the derivatives

By the sum - difference rule of differentiation $\frac{d}{dt}(u\pm v\pm w)=\frac{du}{dt}\pm\frac{dv}{dt}\pm\frac{dw}{dt}$, where $u = t^{4}$, $v = 3\sec t$, $w = 5e^{t}$. So $\frac{ds}{dt}=4t^{3}-3\sec t\tan t + 5e^{t}$.

Answer:

$4t^{3}-3\sec t\tan t + 5e^{t}$