# Math Help - Implicit Differentiation of an equation.

1. ## Implicit Differentiation of an equation.

Guys, I'm having a hard time on understanding the Implicit Differentiation of an equation. Now I'm asking some help to all of you. I will give an equation and KINDLY tell it to me how you got the answer ..Thanks

Assuming that the equation determines a differentiable function f such that y=f(x), find y'

1. 2x^3+x^2y+y^3=1
2.√x+√y=100
3. xy=tan y

Thanks for those who will answer and I would be glad if you can explain to me how you got the answer..

2. ## Re: Implicit Differentiation of an equation.

Originally Posted by coen000
Guys, I'm having a hard time on understanding the Implicit Differentiation of an equation. Now I'm asking some help to all of you. I will give an equation and KINDLY tell it to me how you got the answer ..Thanks

Assuming that the equation determines a differentiable function f such that y=f(x), find y'

1. 2x^3+x^2y+y^3=1
The chain rule: if f(y) is a function of y and y is a function of x, then $\frac{df}{dx}= \frac{df}{dy}\frac{dy}{dx}= \frac{df}dy}y'$. Anytime you differentiate an expression in y, with respect to x, differentiate with respect to y and multiply by y'.

By the chain rule, the deirivative of $x^2y$, with respect to x, is $2xy+ x^2y'$. The derivative of $y^3$, with respect to x, is $3y^2y'$. Of course, the derivative of $2x^3$, with respect to x, is $6x^2$. The derivative of the constant, 1, is, of course, 0. Put those together and solve for y'.

2.√x+√y=100
$\sqrt{x}+ \sqrt{y}= x^{1/2}+ y^{1/2}= 1$
The derivative of $x^{1/2}$, with respect to x, is $\frac{1}{2}x^{-1/2}$ and the derivative of $y^{1/2}$, with respect to x, is [tex]\frac{1}{2}y^{-1/2}y'[tex]
Solve $\frac{1}{2}x^{-1/2}+ \frac{1}{2}y^{-1/2}y'= 0$ for y'.

3. xy=tan y
The derivative of xy, with respect to x, is y+ xy'. The derivative of tan(y), with respect to x, is $sec^2(y)y'$.
Solve $y+ xy'= sec^2(x)y'$ for y'.

Thanks for those who will answer and I would be glad if you can explain to me how you got the answer..