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Math Help - Implicit differentiation.

  1. #1
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    Implicit differentiation.

    I am being asked to find \frac{dy}{dx} by implicit differentiation of the function y=x^{1/n}. I'm really not sure where to begin yet with this problem. I watched a few videos on implicit differentiation but I'm still new to the process. I'm hoping to get some pointers on how to start working this problem.

    Thanks in advance...
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  2. #2
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    Re: Implicit differentiation.

    Hey sepoto.

    In this case you don't need to use implicit differentiation since you have y = f(x).

    But for your information implicit differentiation usually involves differentiating both sides and then putting dy/dx on one side and the rest on the other.
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  3. #3
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    Re: Implicit differentiation.

    Firstly we should know what are implicit functions. in simple words implicit function is a function where in we cannot express one variable it terms of another e.g., xy + x^2y + 3y + 4x = 0; etc.
    In such cases differentiate the function, we get
    y + x dy/dx + 2xy + x^2 dy/dx + 3 dy/dx + 4 = 0
    Now collect all terms containing dy/dx on one side and find the value of dy/dx;
    in the example under consideration we will get
    ( x + x^2 + 3 ) dy/dx = -y-4
    dy/dx = -(y+4)/( x + x^2 + 3 )
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  4. #4
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    Re: Implicit differentiation.

    Raise both sides to the n-th power. So, y^n = (y)^n = (x^{1/n})^n = x^{n/n} = x. Then, take the derivative of both sides:

    \dfrac{d}{dx}(y^n) = \dfrac{d}{dx}(x)

    On the LHS, you need to apply the chain rule since you are differentiating with respect to x. On the RHS, you just get 1.

    \dfrac{d}{dy}(y^n) \dfrac{dy}{dx} = 1

    Now, you can differentiate with respect to y and get:

    ny^{n-1} \dfrac{dy}{dx} = 1

    Solving for \dfrac{dy}{dx}, you have:

    \dfrac{dy}{dx} = \dfrac{1}{ny^{n-1}}

    From your initial equation, you have y = x^{1/n}. Plugging that in, you have:

    \dfrac{dy}{dx} = \dfrac{1}{n(x^{1/n})^{n-1}} = \dfrac{1}{nx^{(n-1)/n}} = \dfrac{1}{n} x^{-(n-1)/n} = \dfrac{1}{n} x^{(1/n-1)}

    This shows that the Power Rule applies even when the exponent of x is a rational number, not just an integer. You won't see a proof that the power rule works when the exponent for x is any real number until you get to logarithmic and exponential differentiation.
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