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Math Help - Implicit Differentiation of an equation.

  1. #1
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    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..
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  2. #2
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    Re: Implicit Differentiation of an equation.

    Quote Originally Posted by coen000 View Post
    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..
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