Inverse Differentiation: Exercise 2

by Batool Akmal

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    00:01 Our next example is asking us to differentiate y equals to 5 sin inverse of 3x. Now, if this was 5 sin of 3x, that would be no problem. But here, we’re going to have to use inverse differentiation or the derivative of it. If we look at y equals to 5 sin inverse of 3x, now usually unless you’re asked to derive the result, you don’t really have to show it. You have the standard result in your formula booklets or in textbooks that would say that if you have y equals to sin inverse of x, the answer to that dy/dx is 1 over √(1 – x²). Worst case scenario, if you ever need to, you can derive this yourselves. You can just do this from scratch like we did earlier on in the lecture.

    00:56 But in this case, we’re just going to use this result. Now, remember that this result is the answer to sin inverse of x. However, we have sin inverse of 3x. What you have here is a function within a function. So you have 3x inside of sin inverse of x. Let’s talk through how we’ll do this, dy/dx.

    01:19 Firstly, 5 is just a constant so that can just stay. The next thing, we have to apply the chain rule.

    01:27 So we’re looking at a function of a function. The chain rule states that we need to differentiate the outside function first just as it is and then we multiply it with the differential of the inside function.

    01:40 Now, the differential of the outside function is this. So sin inverse goes to this.

    01:46 But remember, it’s sin inverse of 3x, not of x. So we can now differentiate sin inverse, just the outside function to 1 over √1– but instead of the x, we’re going to use 3x².

    02:04 Let me just repeat that again. We’re just doing the outside function first. We’ll come to the inside function in a minute. The differential of sin inverse of x is this result here. The differential of sin inverse as a whole if you’re just looking at the outside function of something that is an x, you can just replace that there.

    02:25 But remember that that is just the differential of the outside function. We haven’t finished differentiating sin inverse of 3x. We’ve replaced the x with 3x. Then now, we move to the inside function.

    02:37 So, the differential of 3x is just 3. We’re multiplying this function with 3. We can tidy this up.

    02:44 So at the top, you have 5 times 1 times 3 to just give you 15. At the bottom, you can tidy the square root up a little bit. So you have 1 –. When you square this, you get 9x². That is the answer or the gradient to an inverse function using the standard result. But remember, you are always able to derive the result if you need as well just using the rules that we did earlier in the lecture.

    About the Lecture

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

    Included Quiz Questions

    1. dy/dx=15 / √(1-x²)
    2. dy/dx=15 / √(1+x²)
    3. dy/dx=5 / √(1-x²)
    4. dy/dx= -15 / √(1-x²)
    5. dy/dx= -15 / √(1+x²)
    1. dy/dx = 14x / √( 1-x⁴ )
    2. dy/dx = 14x / √( 1-x² )
    3. dy/dx = 7x / √( 1-x⁴ )
    4. dy/dx = 7 / √( 1-x⁴ )
    5. dy/dx = 14x / √( 1+x² )
    1. dy/dx = -14 / √( 1-x⁴ )
    2. dy/dx = 14 / √( 1-x⁴ )
    3. dy/dx = -14 / √( 1+x⁴ )
    4. dy/dx = 14 / √( 1+x⁴ )
    5. dy/dx = -14x / √( 1+x⁴ )

    Author of lecture Inverse Differentiation: Exercise 2

     Batool Akmal

    Batool Akmal

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