Differentiation of Inverse Functions: Example 2

by Batool Akmal

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    00:01 Now that we’ve done that and we’ve practiced a little bit of changing dx/dy to dy/dx, let’s now move to the more fun parts. We’re going to learn how to differentiate sin inverse of x, cos inverse of x and tan inverse of x. Now, these are standard rules. So, we have set results for these already which you can use but we’ll alternate between deriving the actual result and also just by using the result in some of the questions. But let’s just have a go at firstly deriving the actual differential result by working through this. We’ll learn a little bit about inverse differentiation as we go along. If we look at this question here, we have y equals to sin inverse of x. Now, so far we know that the differential of sin of x is cos of x but we haven’t yet done sin inverse of x. Now, the first step that you can do here is you can obviously change this inverse back to its original. So remember where the sin inverse comes from, it must have come from this side of the equation. So, you would have had sin of y which you can take to the other side to make it into a sin inverse of x. So if I take it back and rewrite this as sin y = x and then I write the x first, so I’m just rewriting this equation. So I write it as x = sin of y. Remember we’re trying to find the gradient, so we’re trying to find dy/dx. The first thing that we could do is rather rearranging it because this is the equation rearranged. We’ve taken it back to basics or to what it should look like.

    01:39 So we’ve taken the sine back. Instead of doing anything else fiddly with this, we can just differentiate here. So we can say let’s differentiate dx with respect to dy. That gives you cos of y.

    01:51 So you’re just differentiating sin y with respect to y which gives you cos y. This doesn’t mean anything, dx/dy at the moment. But you can change it to the gradient by just flipping it. So we want dy/dx, so you’re trying to just change places. You’re swapping it. If you do that here, imagine this is over 1, you’re going to have to do that on the other side as well. So you’re flipping the whole equation.

    02:18 We have dy/dx equals to 1 over cos of y. Now, we’re almost done but the standard result that you usually see in textbooks or in formula books gives this answer in terms of x's and we have cos y.

    02:35 We have dy/dx or the gradient of sin inverse of x here is 1 over cos y. Now, if we want to change this to an answer in terms of x, we’re going to have to use some identities. The identity that we use is this, so sin²y + cos²y = 1. We’ve learned that sin²x + cos²x = 1 or sin²θ + cos²θ = 1. It really doesn’t matter what the angle is. So we’re going to use this identity, rearrange this so we’ll get cos²y = 1 – sin²y.

    03:14 Just to get cos y by itself, this is what we want, you can square root the other side.

    03:20 So, we’ve got cos y = √(1 – sin²y). Let’s replace this now. So we’re going to substitute this in here.

    03:30 We can now say that dy/dx equals to 1 over √(1 – sin²y). Still in terms of y, not in terms of x, if you just look at the very first thing you wrote, so we said that sin y or sin²y in this case is the same as this bit here. So we said that sin y = x therefore this value must be x².

    04:02 Let me just repeat that again. At the start, you said that x was equal to sin y.

    04:07 Here, you can see your sin y but you’ve got two of them. So you got sin² of y. That must be equal to x².

    04:14 You can now change this to 1 over √(1 – x²). If you look at your textbooks or if you look at your formula books, you will often be given y equals to sin inverse of x. Differentiate.

    04:30 So the derivative of that dy/dx is 1 over √(1 –x²). It really depends on the question, what they ask, whether they want you to just go straight to the result which is here or whether they want you to derive the result which is what we’ve done here. But it’s nice to know that just by using inverse differentiation and by using the fact that you can change dx/dy to dy/dx, we can actually derive the actual answer ourselves. So you don’t really need your formula books or textbooks when you can do it all by yourself.

    About the Lecture

    The lecture Differentiation of Inverse Functions: Example 2 by Batool Akmal is from the course Differentiation of Inverse Functions.

    Included Quiz Questions

    1. dy/dx = 1 / √(1 - x²)
    2. dy/dx = 1 / √(1 + x²)
    3. dy/dx = -1 / √(1 - x²)
    4. dy/dx = -1 / √(1 + x²)
    5. dy/dx = √(1 - x²)
    1. x = cos(y) , dx/dy = -sin(y)
    2. x = sin(y) , dx/dy = cos(y)
    3. x = 1/cos(y) , dx/dy = sin(y)/cos²(y)
    4. x = cos(y) , dx/dy = sin(y)
    5. x = cos⁻¹(y) , dx/dy = -1/√(1-y²)
    1. cosΘ = √(1 - sin²Θ)
    2. cosΘ = √(1 + sin²Θ)
    3. cosΘ = 1 - sin²Θ
    4. cosΘ = 1 + sin²Θ
    5. cosΘ = √(1 - sinΘ)

    Author of lecture Differentiation of Inverse Functions: Example 2

     Batool Akmal

    Batool Akmal

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