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 $t$.

$f\left(t\right)=\frac{w}{uv}$

Short answer

The derivative is

$\frac{df}{dt}=\frac{1}{uv}\frac{dw}{dt}-\frac{w}{u^{2}v}\frac{du}{dt}-\frac{w}{u{v}^2}\frac{dv}{dt}$

Step-by-step solution

Step 1. Given Information

The function is

$f\left(t\right)=\frac{w}{uv}$

Step 2. Finding the derivative

The function is

$f\left(t\right)=\frac{w}{uv}$

Differentiate both sides with respect to $t$

$\frac{df}{dt}=\frac{d}{dt}\left(\frac{w}{uv}\right)$

Apply quotient rule of derivative

$\frac{df}{dt}=\frac{uv{\displaystyle \frac{dw}{dt}}-w{\displaystyle \frac{d}{dt}}\left(uv\right)}{{\left(uv\right)}^2}$

Apply product rule of derivative

$\frac{df}{dt}=\frac{uv{\displaystyle \frac{dw}{dt}}-w(v{\displaystyle \frac{du}{dt}}+u{\displaystyle \frac{dv}{dt}})}{{\left(uv\right)}^2}$

Simplify it

$$\begin{gathered} \frac{df}{dt}=\frac{uv{\displaystyle \frac{dw}{dt}}-w(v{\displaystyle \frac{du}{dt}}+u{\displaystyle \frac{dv}{dt}})}{{\left(uv\right)}^2} \\ \frac{df}{dt}=\frac{uv{\displaystyle \frac{dw}{dt}}-wv{\displaystyle \frac{du}{dt}}-wu{\displaystyle \frac{dv}{dt}}}{{\left(uv\right)}^2} \\ \frac{df}{dt}=\frac{uv}{{\left(uv\right)}^2}\frac{dw}{dt}-\frac{wv}{{\left(uv\right)}^2}\frac{du}{dt}-\frac{wu}{{\left(uv\right)}^2}\frac{dv}{dt} \\ \frac{df}{dt}=\frac{1}{uv}\frac{dw}{dt}-\frac{w}{u^{2}v}\frac{du}{dt}-\frac{w}{{uv}^2}\frac{dv}{dt} \end{gathered}$$