Question

find $\frac{ds}{dt}$. s = 4t$^{5}$ - 2 sec t + 3e$^{t}$ $\frac{ds}{dt}=square$

Explanation:

Step1: Differentiate $4t^5$

Using the power - rule $\frac{d}{dt}(at^n)=nat^{n - 1}$, for $a = 4$ and $n = 5$, we have $\frac{d}{dt}(4t^5)=4\times5t^{5 - 1}=20t^4$.

Step2: Differentiate $-2\sec t$

The derivative of $\sec t$ is $\sec t\tan t$, so $\frac{d}{dt}(-2\sec t)=-2\sec t\tan t$.

Step3: Differentiate $3e^t$

The derivative of $e^t$ is $e^t$, so $\frac{d}{dt}(3e^t)=3e^t$.

Step4: Combine the derivatives

By the sum - rule of differentiation $\frac{d}{dt}(u + v+w)=\frac{du}{dt}+\frac{dv}{dt}+\frac{dw}{dt}$, where $u = 4t^5$, $v=-2\sec t$, and $w = 3e^t$. So $\frac{ds}{dt}=20t^4-2\sec t\tan t + 3e^t$.

Answer:

$20t^4-2\sec t\tan t + 3e^t$