Implicit differentiation

[dear2009]

Dear mathhelp participants,


I need help or tips with the following problems



Find d / dx at the point (0, p/2) on the curve implicitly defined by
cos [Pi(x^3)-y] = xy

when I did the work myself I got 3.14. Just curious to know if I did something wrong




Find dy/dx at the point (1, 1) on the curve implicitly defined by
(x^3)(y) -3(x^2) + 4(y^3) = 2


Any help would be nice.

 

[soroban]

Hello, dear2009!

Here's the first one . . .

$\displaystyle \text{Find }\frac{dy}{dx}\text{ at the point }(0,\:\tfrac{\pi}{2}) \text{ on the curve implicitly defined by: }\:\cos (\pi x^3-y) \:=\: xy$

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

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

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

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

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


$\displaystyle \text{At }(0,\:\tfrac{\pi}{2})\!:\quad\frac{dy}{dx} \;=\;\frac{0\cdot\sin(-\frac{\pi}{2}) + \frac{\pi}{2}} {\sin(-\frac{\pi}{2}) - 0} \;=\;\frac{0 + \frac{\pi}{2}}{-1 - 0} \;=\;-\frac{\pi}{2}$

 

[fasteddie65]

Find dy/dx at the point (1, 1) on the curve implicitly defined by
(x^3)(y) -3(x^2) + 4(y^3) = 2

x^2 • dy/dx + 3x^2y - 6x + 12y^2 dy/dx = 0
(x^2 + 12y^2) dy/dx = 6x - 3x^2y
dy/dx = (6x - 3x^2y)/(x^2 + 12y^2) = (6•1 - 3•1^2•1)/(1^2 + 12•1^2) = (6 - 3)/(1 + 12) = 3/13

 

[BigGlenntheHeavy]

Here is the graph of the second one.

[attachment=0:1xgkwk13]like.jpg[/attachment:1xgkwk13]