Implicit differentiation

[asimon2005]

Find dy/dx by implicit differentiation.

14. Y sin(x^2) = x sin(y^2)

18. Tan(x-y) = y/1+x^2

 

[soroban]

Hello, asimon2005!

$\displaystyle \text{Find }\frac{dy}{dx}\text{ by implicit differentiation..}$

. . $\displaystyle 14.\;\;y\cdot\sin(x^2) \:=\: x\cdot\sin(y^2)$

$\displaystyle \text{Product Rule: }\;y\!\cdot\!\cos(x^2)\!\cdot\!2x + \frac{dy}{dx}\!\cdot\!\sin(x^2) \;=\;x\!\cdot\!\cos(y^2)\!\cdot\!2y\!\cdot\!\frac{dy}{dx} + 1\!\cdot\!\sin(y^2)$

$\displaystyle \text{We have: }\:2xy\cos(x^2) + \sin(x^2)\frac{dy}{dx} \;=\; 2xy\cos(y^2)\frac{dy}{dx} + \sin(y^2)$

$\displaystyle \text{Rearrange terms: }\;\sin(x^2)\frac{dy}{dx} - 2xy\cos(y^2)\frac{dy}{dx} \;=\;\sin(y^2) - 2xy\cos(x^2)$

$\displaystyle \text{Factor: }\;\bigg[\sin(x^2) - 2xy\cos(y^2)\bigg]\frac{dy}{dx} \;=\;\sin(y^2) - 2xy\cos(x^2)$


$\displaystyle \text{Therefore: }\;\frac{dy}{dx} \;=\;\frac{\sin(y^2) - 2xy\cos(x^2)}{\sin(x^2) - 2xy\cos(y^2)}$

$\displaystyle 18.\;\;\tan(x-y) \:=\:\frac{y}{1+x^2}$

$\displaystyle \text{We have: }\;(1+x^2)\cdot\tan(x-y) \;=\;y$

$\displaystyle \text{Product Rule: }\;(1 + x^2)\!\cdot\!\sec^2(x-y)\!\cdot\!\left(1 - \frac{dy}{dx}\right) + (2x)\!\cdot\!\tan(x-y) \;=\;\frac{dy}{dx}$

$\displaystyle \text{We have: }\;(1+x^2)\!\cdot\!\sec^2(x-y) - (1+x^2)\!\cdot\!\sec^2(x-y)\frac{dy}{dx} + 2x\!\cdot\!\tan(x-y) \;=\; \frac{dy}{dx}$

$\displaystyle \text{Rearrange terms: }\;\frac{dy}{dx} \;+\; (1+x^2)\sec^2(x-y)\frac{dy}{dx} \;=\;(1+x^2)\sec^2(x-y) \;+\; 2x\tan(x-y)$

$\displaystyle \text{Factor: }\;\bigg[1 + (1+x^2)\sec^2(x-y)\bigg]\frac{dy}{dx} \;=\;(1+x^2)\sec^2(x-y) + 2xy\tan(x-y)$


$\displaystyle \text{Therefore: }\;\frac{dy}{dx} \;=\;\frac{(1+x^2)\sec^2(x-y) + 2xy\tan(x-y)}{1 + (1+x^2)\sec^2(x-y)}$

 

[asimon2005]

Gracias.