How do you find d/dt given y=t^3 - 3t and x=t^2?

Oct 5, 2016

It depends upon what you intended by $\frac{d}{\mathrm{dt}}$

See possible interpretations below:

Explanation:

Option 1: You wanted $\textcolor{b l a c k}{\frac{\mathrm{dc} o l \mathmr{and} \left(red\right) y}{\mathrm{dt}}}$
$\frac{\mathrm{dy}}{\mathrm{dt}} = 3 {t}^{2} - 3$ (using the exponent rule)

Option 2: You wanted $\textcolor{b l a c k}{\frac{\mathrm{dc} o l \mathmr{and} \left(red\right) x}{\mathrm{dt}}}$
$\frac{\mathrm{dx}}{\mathrm{dt}} = 2 t$ (using the exponent rule)

Option 3: You wanted color(black)((dy)/(dcolor(red)x)
...definitely not what was asked for, but a possibly intended requirement.
If $y = {t}^{3} - 3 t$ and $x = {t}^{2}$
then $t = {x}^{\frac{1}{2}}$ and ${t}^{3} = {x}^{\frac{3}{2}}$
so
$y = {x}^{\frac{3}{2}} - 3 {x}^{\frac{1}{2}}$
and
$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{3}{2} {x}^{\frac{1}{2}} - \frac{3}{2} {x}^{- \frac{1}{2}} = \frac{3 \sqrt{x}}{2} - \frac{3}{2 \sqrt{x}}$