# How do you differentiate the following parametric equation:  x(t)=t^2+tcos2t, y(t)=tsint ?

Jul 8, 2017

$x ' \left(t\right) = 2 t - 4 \sin \left(2 t\right)$

$y ' \left(t\right) = \sin t + t \cos t$

#### Explanation:

We're asked to find the derivative of a set of parametric equations.

We differentiate each individual equation as normal:

x(t)

We can differentiate term by term, and factor out the constant,$2$ in the second term:

$x ' \left(t\right) = \frac{d}{\mathrm{dt}} \left[{x}^{2}\right] + 2 \frac{d}{\mathrm{dt}} \left[\cos \left(2 t\right)\right]$

Use the power rule:

$\frac{d}{\mathrm{dt}} \left[{t}^{n}\right] = n {t}^{n - 1}$

where $n = 2$:

$= \textcolor{red}{2 t} + 2 \frac{d}{\mathrm{dt}} \left[\cos \left(2 t\right)\right]$

Using the chain rule:

$\frac{d}{\mathrm{dt}} \left[\cos \left(2 t\right)\right] = \frac{\mathrm{dc} o s u}{\mathrm{du}} \frac{\mathrm{du}}{\mathrm{dt}}$

where $u = 2 t$ and $\frac{d}{\mathrm{du}} \left[\cos u\right] = - \sin u$:

$= 2 t + 2 \frac{d}{\mathrm{dt}} \left[2 t\right] \cdot - \sin \left(2 t\right)$

Simplifying:

$= 2 t - 2 \frac{d}{\mathrm{dt}} \left[2 t\right] \sin \left(2 t\right)$

Factor out the constant, $2$:

$= 2 t - 4 \frac{d}{\mathrm{dt}} \left[t\right] \sin \left(2 t\right)$

The derivative of $x$ is $1$

color(red)(= 2t - 4sin(2t)

y(t)

Use the product rule:

$\frac{d}{\mathrm{dt}} \left[u v\right] = v \frac{\mathrm{du}}{\mathrm{dt}} + u \frac{\mathrm{dv}}{\mathrm{dt}}$

where $u = t$ and $v = \sin t$:

$y ' \left(t\right) = t \frac{d}{\mathrm{dt}} \left[\sin t\right] + \frac{d}{\mathrm{dt}} \left[t\right] \sin t$

The derivative of $\sin t$ is $\cos t$:

$= \frac{d}{\mathrm{dt}} \left[t\right] \sin t + t \cos t$

The derivative of $t$ is $1$:

color(blue)(= sint + tcost