Implicit Differentiation: Exercise 3

by Batool Akmal

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    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. We are looking at differentiating cos y plus 2x to the 5 equals to 10. 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 take with differentiating this implicitly. 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. Remember that cos goes to sine but this is going to go to minus sine y. 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. 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. And then finally, 10x to the 4 over sine y equals to dy/dx. 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. ...

    About the Lecture

    The lecture Implicit Differentiation: Exercise 3 by Batool Akmal is from the course Implicit Differentiation.

    Author of lecture Implicit Differentiation: Exercise 3

     Batool Akmal

    Batool Akmal

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