Question

Given that $u=u\left(t\right),v=v\left(t\right),$and $w=w\left(t\right)$are functions of $t$and that $k$is a constant, calculate the derivative $\frac{df}{dt}$of each function $f\left(t\right)$. Your answers may involve $u,v,w,$$\frac{du}{dt},\frac{dv}{dt},\frac{dw}{dt}$, $k$, and/or $t$.

$f\left(t\right)=\frac{k}{u^{2}w}$

Short answer

The derivative is

$\frac{df}{dt}=-\frac{2k}{u^{3}w}\frac{du}{dt}-\frac{k}{u^2{w}^2}\frac{dw}{dt}$

Step-by-step solution

Step 1. Given Information

The function is

$f\left(t\right)=\frac{k}{u^{2}w}$

Step 2. Finding the derivative

The function is

$f\left(t\right)=\frac{k}{u^{2}w}$

Differentiate both sides with respect to $t$

$\frac{df}{dt}=\frac{d}{dt}\left(\frac{k}{u^{2}w}\right)$

Take constant out of derivative

$$\begin{gathered} \frac{df}{dt}=k\frac{d}{dt}\left(\frac{1}{u^{2}w}\right) \\ \frac{df}{dt}=k\frac{d}{dt}\left(u^{-2}{w}^{-1}\right) \end{gathered}$$

Apply product rule of derivative

$\frac{df}{dt}=k\left[w^{-1}\frac{d}{dt}\right(u^{-2})+u^{-2}\frac{d}{dt}(w^{-1}\left)\right]$

Apply power and chain rule of derivative

$\frac{df}{dt}=k\left[w^{-1}\right(-2u^{-3}\frac{du}{dt})+u^{-2}(-1w^{-2}\frac{dw}{dt}\left)\right]$

Simplify it

$$\begin{gathered} \frac{df}{dt}=k\left[w^{-1}\right(-2u^{-3}\frac{du}{dt})+u^{-2}(-1w^{-2}\frac{dw}{dt}\left)\right] \\ \frac{df}{dt}=k[-\frac{2}{u^{3}w}\frac{du}{dt}-\frac{1}{u^2{w}^2}\frac{dw}{dt})] \\ \frac{df}{dt}=-\frac{2k}{u^{3}w}\frac{du}{dt}-\frac{k}{u^2{w}^2}\frac{dw}{dt} \end{gathered}$$