Implicit Differentiation

[dear2009]

Dear MathHelp Participants,


Find dy/dx at the point (0, p/2) on the curve implicitly defined by
cos (pix^3 - y) = xy


This is what I did so far
-sin (3pix^2) = y

but I know I am missing some stuff, could somebody show how to properly setup this derivative. Thanks in advance.

 

[soroban]

Hello, dear2009!

$\displaystyle \text{Find }\frac{dy}{dx}\text{ at the point }\left(0,\:\frac{\pi}{2}\right)\text{ on the curve implicitly defined by:}$

. . $\displaystyle \cos (\pi x^3 - y) \:=\: xy$


$\displaystyle \text{This is what I did so far: }\;-\sin (3\pi x^2) \:=\: y$

but I know I am missing some stuff . . .
. . . Yeah, like ALL of implicit differentiation

$\displaystyle \text{Differentiate implicitly: }\;-\sin(\pi x^3-y)\cdot\!\left(3\pi x^2 - \frac{dy}{dx}\right) \;=\;x\!\cdot\!\frac{dy}{dx} + 1\!\cdot\!y$

. . $\displaystyle -3\pi x^2\sin(\pi x^3-y) + \sin(\pi x^3-y)\!\cdot\!\frac{dy}{dx} \;=\;x\!\cdot\!\frac{dy}{dx} + y$

. . $\displaystyle \sin(\pi x^3-y)\!\cdot\!\frac{dy}{dx} - x\!\cdot\!\frac{dy}{dx} \;=\;3\pi x^2\sin(\pi x^3-y) + y$

. . $\displaystyle \bigg[\sin(\pi x^3-y) - x\bigg]\!\cdot\!\frac{dy}{dx} \;=\;3\pi x^2\sin(\pi x^3-y) + y$

. . $\displaystyle \boxed{\frac{dy}{dx} \;=\;\frac{3\pi x^2\sin(\pi x^3-y) + y}{\sin(\pi x^3 - y) - x}}$



$\displaystyle \text{At }\left(0,\:\frac{\pi}{2}\right)\!:\;\;\frac{dy}{dx} \;=\;\frac{\overbrace{3\pi(0^2)\sin(0- \tfrac{\pi}{2})}^{\text{This is 0}} + \dfrac{\pi}{2}} {\underbrace{\sin(0 - \tfrac{\pi}{2})}_{\text{This is -1}} - 0} \;=\;\frac{\frac{\pi}{2}}{\text{-}1} \;=\;-\frac{\pi}{2}$