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Chain and Product Rule – Implicit Differentiation

by Batool Akmal
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    So after those 2 examples, we can now summarize what we?ve just learned. Remember what we?re saying, if you have a function of x in terms of y and you?re trying to differentiate y with respect to x, you can do that but you'll have to differentiate y with respect to y because that?s the only way we can differentiate y. So we can write that as d/dy of y and you must remember to multiply it with the factor dy/dx to make up for the extra factor that you have just multiplied in. You might find doing this numerically a lot easier than this actual definition so we?ll keep practicing. Let?s move on to a more difficult example. Let?s look at our next example. You can see already that this has a mixture of x?s and y?s, so hopefully you are recognizing that this is an implicit equation. Let?s write this out and talk about each term separately. We have 2x cubed plus xy equals to y squared plus 10. Start to look at each term and thinking about what we?re going to do. We?ve already decided that this is an implicit differentiation question so we?re not going to attempt to rearrange for y, we?re just going to differentiate it all in one go. Now, keep your objective clear. We are trying to find dy/dx, that is the gradient. So every time you are doing these questions, remember that you want to find dy/dx not dx/dy or no other variations of it. You have to get dy/dx by itself and then get your answer. So let?s start to look at each individual term. This we can differentiate with respect to x, this we can differentiate with respect to y but we?re to multiply this with dy/dx. Here is a...

    About the Lecture

    The lecture Chain and Product Rule – Implicit Differentiation by Batool Akmal is from the course Implicit Differentiation.


    Included Quiz Questions

    1. f(x) g'(x) + g(x) f'(x)
    2. f(x) g(x) + f(x) g(x)
    3. f(x) + g'(x)
    4. g'(x) f'(x) + f(x) g(x)

    Author of lecture Chain and Product Rule – Implicit Differentiation

     Batool Akmal

    Batool Akmal


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