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Differentiation: Exercise 2 – Calculus Methods

by Batool Akmal
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    So moving on from the previous question where we were just asked to find the general gradient. This questions move up one step further, it's asking you the gradient at x equals to 5. So all that means is that we do the same process like in the general gradient and then we substitute 5 into the equation. So there is no need to change any roots or any indices here. So we have y equals 10x to the 5 minus 2x plus 7. Just rewrite it again so we can see where the powers are going. dy by dx. Bring the power down. So we have 5 times 10x, 5 minus 1 minus 1. And then we've got 2x, 1 minus 1. And anything as a constant just disappears. This gives me 50x to the 4 minus 2. We are now looking at the gradient at x equals to 5. So we've got 50x to the 5. Let's just x to the 4 I mean. Let's just say what we're doing, so dy by dx at x = 5 - 2. You'd have to work out what 5 to the power of 4 is. So we know where 5 squared is, means 25. You multiply that with another 5 to give you a 125. You multiply that with another 5. So again remember that we are not quite using calculators here. So you can't just work it out on the side. And it's important that you do. Because there's no shame in working things on the side as long as you don't get it wrong. So you have 25. 5 times 2 is 10 and plus 2 gives you 12. And then we've got 5 times 1. So we got 625. So here we can rewrite this as 50...

    About the Lecture

    The lecture Differentiation: Exercise 2 – Calculus Methods by Batool Akmal is from the course Calculus Methods: Differentiation.


    Author of lecture Differentiation: Exercise 2 – Calculus Methods

     Batool Akmal

    Batool Akmal


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