Edin's question via email about implicit differentiation

[Prove It]

A curve has equation $\displaystyle \begin{align*} y^3 + y + x\,y^2 = 10 + 4\sin{(x)} \end{align*}$.

(a) Determine y' at the point (0, 2)

(b) Determine y'' at the point (0,2)

(a) Differentiate both sides of the equation with respect to x:

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

so at (0, 2) we have

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{4\cos{(0)} - 2^2}{3 \cdot 2^2 + 1 + 2 \cdot 0 \cdot 2} \\ &= \frac{0}{13} \\ &= 0 \end{align*}$(b) Differentiate both sides of the resulting equation with respect to x again...

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[ \left( 3\,y^2 + 1 + 2\,x\,y \right) \frac{\mathrm{d}y}{\mathrm{d}x} \right] &= \frac{\mathrm{d}}{\mathrm{d}x} \left[ 4\cos{(x)} - y^2 \right] \\ \left( 3\,y^2 + 1 + 2\,x\,y \right) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left( 6\,y\,\frac{\mathrm{d}y}{\mathrm{d}x} + 2\,y + 2\,x\,\frac{\mathrm{d}y}{\mathrm{d}x} \right) \frac{\mathrm{d}y}{\mathrm{d}x} &= -4\sin{(x)} - 2\,y\,\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}$

so at (0, 2) where $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \end{align*}$ we have

$\displaystyle \begin{align*} \left( 3\cdot 2^2 + 1 + 2\cdot 0 \cdot 2 \right) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left( 6 \cdot 2 \cdot 0 + 2\cdot 2 + 2 \cdot 0 \cdot 0 \right) \cdot 0 &= -4\sin{(0)} - 2\cdot 2 \cdot 0 \\ \left( 12 + 1 + 0 \right) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 0 &= 0 + 0 \\ 13\,\frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= 0 \\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= 0 \end{align*}$

 

[chwala]

(a) Differentiate both sides of the equation with respect to x:

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

so at (0, 2) we have

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{4\cos{(0)} - 2^2}{3 \cdot 2^2 + 1 + 2 \cdot 0 \cdot 2} \\ &= \frac{0}{13} \\ &= 0 \end{align*}$(b) Differentiate both sides of the resulting equation with respect to x again...

$\displaystyle \begin{align*} \frac{\mathrm{d}}{\mathrm{d}x} \left[ \left( 3\,y^2 + 1 + 2\,x\,y \right) \frac{\mathrm{d}y}{\mathrm{d}x} \right] &= \frac{\mathrm{d}}{\mathrm{d}x} \left[ 4\cos{(x)} - y^2 \right] \\ \left( 3\,y^2 + 1 + 2\,x\,y \right) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left( 6\,y\,\frac{\mathrm{d}y}{\mathrm{d}x} + 2\,y + 2\,x\,\frac{\mathrm{d}y}{\mathrm{d}x} \right) \frac{\mathrm{d}y}{\mathrm{d}x} &= -4\sin{(x)} - 2\,y\,\frac{\mathrm{d}y}{\mathrm{d}x} \end{align*}$

so at (0, 2) where $\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} = 0 \end{align*}$ we have

$\displaystyle \begin{align*} \left( 3\cdot 2^2 + 1 + 2\cdot 0 \cdot 2 \right) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + \left( 6 \cdot 2 \cdot 0 + 2\cdot 2 + 2 \cdot 0 \cdot 0 \right) \cdot 0 &= -4\sin{(0)} - 2\cdot 2 \cdot 0 \\ \left( 12 + 1 + 0 \right) \frac{\mathrm{d}^2y}{\mathrm{d}x^2} + 0 &= 0 + 0 \\ 13\,\frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= 0 \\ \frac{\mathrm{d}^2y}{\mathrm{d}x^2} &= 0 \end{align*}$

Correct! Implicit differentiation was applied well.