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Differentiation of Trigonometric Functions: Exercise 2

by Batool Akmal
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    Our next example is now asking us to differentiate 1 over sin x. Now think if you can remember what 1 over sin x was called. It?s cosec of x. But we like to deal with things written in terms of sin, cos, and tan, so rather than overcomplicating it, let?s just start differentiating this. So I have y equals to 1 over sin x. If you identified or recognized that this is the same as cosec, that?s brilliant, you?re obviously learning your identities. But in this case, we can see that we have a number at the top and a number at the bottom. You can take you entire sin x up and then just treat it like the chain rule. So by that I mean you could do this, sin x to the minus 1 and then it?s very similar to the previous example, so you could use it as the chain rule. But the idea is to practice the quotient rule so I won?t do that. We?ll just rewrite this as the quotient of u and v, so I?ll say the top value is u and the bottom value is v. The quotient rule, dy/dx is vdudx minus udvdx over v squared. So we separate it out and then we put it into the formula. So, u equals to 1, v equals to sin of x, u dashed is zero and v dashed is cos of x. Go straight into the formula, we?re using vdudx minus that so dy/dx is just going to be sin of x multiplied by zero minus 1 times cos of x and then you divide it by the squared which is just sin x squared or sin squared x. Sin multiplied by zero just gives you zero, that just goes away, leaving...

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


    Author of lecture Differentiation of Trigonometric Functions: Exercise 2

     Batool Akmal

    Batool Akmal


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