Q 1. TB - Calculus

Question

Using the chain rule: Given that $r=r\left(t\right), s=s\left(t\right),$ and $u=u\left(t\right)$ are functions of $t$ and that $c$ and $k$ are constants, find each of the following derivatives.

$$\begin{gathered} \left(i\right)\frac{d}{dt}\left(\pi u^2\right) \left(ii\right)\frac{d}{dt}\left(3r+2s\right) \left(iii\right)\frac{d}{dt}\left(cu+rs\right) \\ \left(iv\right)\frac{d}{dt}\left(k+u^3\right) \left(v\right)\frac{d}{dt}\left(c{r}^{2}u\right) \left(vi\right)\frac{d}{dt}\left(\frac{c+s}{k+u}\right) \end{gathered}$$

Step-by-Step Solution

Answer

The derivative of $$\begin{gathered} \left(i\right)\frac{d}{dt}\left(\pi u^2\right)=2\pi u\left[u^{\text{'}}\left(t\right)\right] \left(ii\right)\frac{d}{dt}\left(3r+2s\right)=3\left[r^{\text{'}}\left(t\right)\right]+2\left[s^{\text{'}}\left(t\right)\right] \left(iii\right)\frac{d}{dt}\left(cu+rs\right)=c\left[u^{\text{'}}\left(t\right)\right]+r\left[s^{\text{'}}\left(t\right)\right]+s\left[r^{\text{'}}\left(t\right)\right] \\ \left(iv\right)\frac{d}{dt}\left(k+u^3\right)=3u^2\left[u^{\text{'}}\left(t\right)\right] \left(v\right)\frac{d}{dt}\left(c{r}^{2}u\right)=c\left\{r^2\left[u^{\text{'}}\left(t\right)\right]+2ur\left[r^{\text{'}}\left(t\right)\right]\right\} \left(vi\right)\frac{d}{dt}\left(\frac{c+s}{k+u}\right)=\frac{s^{\text{'}}\left(t\right)\left(k+u\right)-u^{\text{'}}\left(t\right)\left(c+s\right)}{{\left(k+u\right)}^2} \end{gathered}$$

Step 1. Definition

In calculus, the chain rule is a formula that expresses the derivative of the composition of two differentiable functions $f$ and $g$ in terms of the derivatives of $f$ and $g$.

Step 2. Explanation and solution

$\left(i\right)\frac{d}{dt}\left(\pi u^2\right)=\pi \frac{d}{dt}\left(u^2\right)=2\pi u\left[u^{\text{'}}\left(t\right)\right]$

$\left(ii\right) \frac{d}{dt}\left(3r+2s\right)=3\frac{d}{dt}\left(r\right)+2\frac{d}{dt}\left(s\right)=3\left[r^{\text{'}}\left(t\right)\right]+2\left[s^{\text{'}}\left(t\right)\right]$

$\left(iii\right) \frac{d}{dt}\left(cu+rs\right)=c\frac{d}{dt}\left(u\right)+r\frac{d}{dt}\left(s\right)+s\frac{d}{dt}\left(r\right)=c\left[u^{\text{'}}\left(t\right)\right]+r\left[s^{\text{'}}\left(t\right)\right]+s\left[r^{\text{'}}\left(t\right)\right]$

$\left(iv\right) \frac{d}{dt}\left(k+u^3\right)=\frac{d}{dt}\left(k\right)+\frac{d}{dt}\left(u^3\right)=3u^2\left[u^{\text{'}}\left(t\right)\right]$

$\left(v\right) \frac{d}{dt}\left(c{r}^{2}u\right)=c\left[r^2\frac{d}{dt}\left(u\right)+u\frac{d}{dt}\left(r^2\right)\right]=c\left\{r^2\left[u^{\text{'}}\left(t\right)\right]+2ur\left[r^{\text{'}}\left(t\right)\right]\right\}$

$\left(vi\right) \frac{d}{dt}\left(\frac{c+s}{k+u}\right)=\frac{\left(k+u\right){\displaystyle \frac{d}{dt}}\left(c+s\right)-\left(c+s\right){\displaystyle \frac{d}{dt}}\left(k+u\right)}{{\left(k+u\right)}^2}=\frac{s^{\text{'}}\left(t\right)\left(k+u\right)-u^{\text{'}}\left(t\right)\left(c+s\right)}{{\left(k+u\right)}^2}$