# Implicit Differentiation: Exercise 2

by Batool Akmal

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00:01 Our next question, you can see that we've got a cos term here, so, we're going to be dealing with a little bit of trigonometry.

00:07 We are looking at differentiating cos y plus 2x to the 5 equals to 10.

00:13 If you try to rearrange this for y you're going to end up with some term with cos inverse of a term which again has going to get messy, so let's just stick with differentiating this implicitly.

00:26 Each term we have to differentiate, so, we're going to start with cos y, remember firstly you're attempting to differentiate a y term and secondly this also has trig, so, you need to remember if you're differentiating cos y with respects to y.

00:43 Remember that cos goes to sine but this is going to go to minus sine y.

00:48 Apologies, coz goes to minus sine, so cos y becomes minus sine y and then you multiply it with the factor of dy/dx because it was a y term plus 2x to the 5, will differentiate to 10x to the 4, when you bring the power down so you times it and decreased the power by 1 equals to 10, which is a constant that just goes to zero.

01:11 All we have to do now is rearrange this equation, so, if I leave the 10x to the 4 here and move the minus sine y to the other side, that becomes positive sine y, dy/dx.

01:23 And then finally, 10x to the 4 over sine y equals to dy/dx.

01:31 So that's your gradient here, so you can substitute numbers and once again to find what the particular gradient at different points will be.

The lecture Implicit Differentiation: Exercise 2 by Batool Akmal is from the course Implicit Differentiation.

### Included Quiz Questions

1. dy/dx = (2 + y) / (7 - x)
2. dy/dx = (2 + y) / 7
3. dy/dx = (2 + y) / (7 + x)
4. dy/dx = 2 / (7 - x)
5. dy/dx = (2 - y) / (7x)

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