# Kamal's Questions via email about Implicit Differentiation

#### Prove It

##### Well-known member
MHB Math Helper

Since we have this relationship between x and y, as the two sides are equal, so are their derivatives. We just have to remember that as y is a function of x, any function of y is also a function of x, with the inner function "y" composed inside whatever is being told to do to the y. So to differentiate these parts the Chain Rule would be needed. All other rules like the product and quotient rules will still apply as well. Anyway, differentiating both sides with respect to x gives

\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x}\,\left( x\,y + x^2 \right) &= \frac{\mathrm{d}}{\mathrm{d}x}\,\left( y^2 \right) \\ x\,\frac{\mathrm{d}y}{\mathrm{d}x} + 1\,y + 2\,x &= \frac{\mathrm{d}y}{\mathrm{d}x}\,\frac{\mathrm{d}}{\mathrm{d}y}\,\left( y^2 \right) \\ x\,\frac{\mathrm{d}y}{\mathrm{d}x} + y + 2\,x &= \frac{\mathrm{d}y}{\mathrm{d}x}\,\left( 2\,y \right) \\ y + 2\,x &= 2\,y\,\frac{\mathrm{d}y}{\mathrm{d}x} - x\,\frac{\mathrm{d}y}{\mathrm{d}x} \\ y + 2\,x &= \left( 2\,y - x \right) \,\frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{y + 2\,x}{2\,y - x} \end{align*}

This will be important for later. Going back a step and differentiating both sides with respect to x again we have

\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \,\left( y + 2\,x \right) &= \frac{\mathrm{d}}{\mathrm{d}x}\,\left[ \left( 2\,y - x \right) \,\frac{\mathrm{d}y}{\mathrm{d}x} \right] \\ \frac{\mathrm{d}y}{\mathrm{d}x} + 2 &= \left( 2\,y - x \right) \,\frac{\mathrm{d}^2\,y}{\mathrm{d}x^2} + \left( 2\,\frac{\mathrm{d}y}{\mathrm{d}x} - 1 \right) \,\frac{\mathrm{d}y}{\mathrm{d}x} \\ \frac{\mathrm{d}y}{\mathrm{d}x} + 2 &= \left( 2\,y - x \right) \,\frac{\mathrm{d}^2\,y}{\mathrm{d}x^2} + 2\,\left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 - \frac{\mathrm{d}y}{\mathrm{d}x} \\ \left( 2\,y - x \right) \,\frac{\mathrm{d}^2\,y}{\mathrm{d}x^2} &= 2 + 2\,\frac{\mathrm{d}y}{\mathrm{d}x} - 2\,\left( \frac{\mathrm{d}y}{\mathrm{d}x}\right) ^2 \\ \left( 2\,y - x \right) \,\frac{\mathrm{d}^2\,y}{\mathrm{d}x^2} &= 2\,\left[ 1 + \frac{\mathrm{d}y}{\mathrm{d}x} - \left( \frac{\mathrm{d}y}{\mathrm{d}x} \right) ^2 \right] \end{align*}

and since we already found that \displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{y + 2\,x}{2\,y - x } \end{align*} that means

\displaystyle \begin{align*} \left( 2\,y - x \right) \,\frac{\mathrm{d}^2\,y}{\mathrm{d}x^2} &= 2 \, \left[ 1 + \frac{y + 2\,x}{2\,y - x} - \left( \frac{y + 2\,x}{2\,y - x } \right) ^2 \right] \\ \left( 2\,y - x \right) \,\frac{\mathrm{d}^2\,y}{\mathrm{d}x^2} &= 2\,\left[ \frac{\left( 2\,y - x \right) ^2 + \left( 2\,y - x \right) \left( y + 2\,x \right) - \left( y + 2\,x \right) ^2}{\left( 2\,y - x \right) ^2 } \right] \\ \left( 2\,y - x \right) \,\frac{\mathrm{d}^2\,y}{\mathrm{d}x^2} &= 2\,\left[ \frac{4\,y^2 - 4\,x\,y + x^2 + 2\,y^2 + 4\,x\,y - x\,y - 2\,x^2 - y^2 - 4\,x\,y - 4\,x^2}{\left( 2\,y - x \right) ^2} \right] \\ \left( 2\,y - x \right) \,\frac{\mathrm{d}^2\,y}{\mathrm{d}x^2} &= 2\,\left[ \frac{5\,y^2 - 5\,x\,y - 5\,x^2 }{\left( 2\,y - x \right) ^2} \right] \\ \frac{\mathrm{d}^2\,y}{\mathrm{d}x^2} &= \frac{10\,\left( y^2 - x\,y - x^2 \right) }{\left( 2\,y - x \right) ^3} \end{align*}

Here x is a function of t, so again, to differentiate any x terms, we must use the Chain Rule.

\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t} \,\left( x^3 + x\,t^{-1} \right) &= \frac{\mathrm{d}}{\mathrm{d}t} \,\left[ t^2 \left( 1 + x^2 \right) \right] \\ \frac{\mathrm{d}x}{\mathrm{d}t} \, \frac{\mathrm{d}}{\mathrm{d}x} \, \left( x^3 \right) + \frac{\mathrm{d}x}{\mathrm{d}t} \, t^{-1} - x \, t^{-2} &= 2 \, t \, \left( 1 + x^2 \right) + t^2 \, \frac{ \mathrm{d}x }{ \mathrm{d}t } \, \frac{ \mathrm{d} }{\mathrm{d}x} \, \left( 1 + x^2 \right) \\ \frac{\mathrm{d}x}{\mathrm{d}t}\,\left( 3\,x^2 \right) + \frac{\mathrm{d}x}{\mathrm{d}t}\,t^{-1} - x \,t^{-2} &= 2\,t\,\left( 1 + x^2 \right) + t^2\,\frac{\mathrm{d}x}{\mathrm{d}t} \,\left( 2\,x \right) \\ 3\,x^2\,\frac{\mathrm{d}x}{\mathrm{d}t} + t^{-1}\,\frac{\mathrm{d}x}{\mathrm{d}t} - 2\,x\,t^2 \,\frac{\mathrm{d}x}{\mathrm{d}t} &= 2\,t + 2\,x^2\,t + x\,t^{-2} \\ t^2\,\left( 3\,x^2 + t^{-1} - 2\,x\,t^2 \right) \,\frac{\mathrm{d}x}{\mathrm{d}t} &= t^2\,\left( 2\,t + 2\,x^2\,t + x\,t^{-2} \right) \\ \left( 3\,x^2\,t^2 + t - 2\,x\,t^4 \right) \,\frac{\mathrm{d}x}{\mathrm{d}t} &= 2\,t^3 + 2\,x^2\,t^3 + x \\ \frac{\mathrm{d}x}{\mathrm{d}t} &= \frac{2\,t^3 + 2\,x^2\,t^3 + x}{3\,x^2\,t^2 + t - 2\,x\,t^4} \end{align*}