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Differentiation of Trigonometric Functions: Exercise 4

by Batool Akmal
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    So the last question, which is a little bit of an algebraic challenge, is 2 functions dividing each other but this time there's no numbers, we?re just dealing with letters. The letters are constant so a and b could be any numbers and we?re just deriving a general formula or a general result. So, we have y equals to a plus b sin of x over a minus b sin of x. Look at your function. What is happening? Two functions are dividing. When you are to differentiate something that?s dividing, you will use the quotient rule. So, these things should be coming more naturally to us. This is our u, this is our v. Our quotient rule is this, dy/dx is vdudx minus udvdx over v squared. Let?s break it down into its u?s and v?s. So u equals to a plus b sin x, v equals to a minus b sin x. Differentiate it, u dashed. Now, here?s the thing, a is any constant, it?s any number so it will just disappear so don?t just keep it there because it?s a letter, a is a number so it disappears, b is a constant so you can imagine it?s 5 sin x and you differentiate it. So b stays and sin x differentiates to cos x. It?s not a function of a function, no chain rule, just straight forward, differentiate sin x to cos x. Similarly for v dashed, so let?s just stay with the same notation rather than dv/dx. The a disappears and then b or the sin x goes to cos x, so this becomes minus b cos of x. Okay, let?s put these all into our formula for our quotient rule, vdudx minus udvdx. So we can now say that dy/dx is b cos x a...

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    The lecture Differentiation of Trigonometric Functions: Exercise 4 by Batool Akmal is from the course Differentiation of Trigonometric Functions.


    Author of lecture Differentiation of Trigonometric Functions: Exercise 4

     Batool Akmal

    Batool Akmal


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