Derivative of sin(x)

by Batool Akmal

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    00:01 So let's get straight into it now, we are now going to derive the answer for the gradient of sin x.

    00:08 We already know the answer to this, I know.

    00:10 So we've discussed already that sin x differentiate the cos of x but we're going to prove it to ourselves, we're gonna develop our algebraic, our proving, our analysis skills a little bit more and we're hopefully going to be excited when we see the answer.

    00:24 So back to the very first lecture of differentiation from first principles, let me just remind you every time we're faced by a function that we want to prove, we can always go back to the definition of the derivative.

    00:36 So if I remind you, if you have, dy/dx of a function, you say limit delta x tends to zero of f of x plus delta x minus f of x over delta x.

    00:55 We are now trying to prove sin of x so we want the derivative of sin of x as the limit of delta x tends to zero. Okay, so instead of using f, we'll now use s, sin of x plus delta x minus the original sin x all over delta x.

    01:20 Right, we can start using our addition law, so if you look at this function here, this is basically sin a plus b. So you can compare this with sin a plus b that we looked up previously, so your a is your x and b is your delta x.

    01:41 So we can use that identity to expand this.

    01:45 We said that sin a plus b is sin a cos b plus sin b cos a but in this case we'll obviously do this with x's and deltas.

    01:54 So we got sin x cos delta x plus sin delta x cos x, so that's this part expanded.

    02:05 Don't forget to subtract it with sin x all divided by delta x.

    02:12 And also don't forget that we still need to consider that the limit of delta x tends to zero.

    02:17 Okay the next thing we can do is we can combine our sin x terms so this term has a sin x and this term has a sin x and combine our cos x terms which is this one.

    02:29 So if we put them together, if I rewrite this one's again, so it's a bit clearer, I have sin x cos delta x and I'm gonna write sin x next to it so this term just change its position and then I have sin delta x cos x which is this term here by itself all over delta x.

    02:50 Okay, this is what I've done this so that I can take a common factor out of these two terms of sin x.

    02:58 So you can see they both have sin x. So I can take sin x out, leaving me with cos delta x minus 1 over delta x and then in this term I can take a common factor of cos x so we'll say cos x leaving us with sin delta x over delta x and remember that for each of them we're still applying that the limit of delta x tends to zero and the limit of delta x tends to zero. Okay, we're almost done now, so sin x is the limit of delta x tends to zero isn't really affected so that change, stays as sin x.

    03:44 We have cos delta x, so this value here you have to think now what happens to cos as it approaches zero. So you can either use a circle or a calculator, let me just remind you what the cos graph looks like.

    03:59 Like so, this is zero, so add zero the cos graph is at 1.

    04:04 So as this graph approaches zero, you can say that this value here approaches 1.

    04:11 So at the top you have 1 minus 1 which gives you zero divided by anything is zero.

    04:17 So you have sin x that's been multiplied by zero.

    04:20 Okay, the next of it is cos x, obviously is an affected by the limit because there's no delta x here.

    04:28 This is our proof that we just did earlier, so we said that sin x over x as the limit of x goes to zero goes to 1 but instead we're now using delta x, delta x and the limit of delta x, same thing. So we know that this entire thing approaches 1.

    04:45 So we can now multiply this with 1 and finally our answer or the final answer if we just write it here to make it a little bit more dramatic, we can say that dy/dx as the limit of delta x times zero is simply cos x because that term goes away. And so, worst case scenario if you ever can't remember the derivative of sin x, you know the whole methods to actually derive it.

    05:09 Hopefully, you don't have to do this, this can just be to prove something but you can learn that sin x differentiate to cos x and the differential of sin x is simply cos x.

    About the Lecture

    The lecture Derivative of sin(x) by Batool Akmal is from the course Differentiation of Trigonometric Functions.

    Included Quiz Questions

    1. cos(x)
    2. -sin(x)
    3. -cos(x)
    4. tan(x)
    5. cot(x)
    1. sin(x)cos(y) + cos(x)sin(y)
    2. sin(x)cos(y) - cos(x)sin(y)
    3. cos(x)cos(y) + sin(x)sin(y)
    4. cos(x)cos(y) - sin(x)sin(y)
    5. 2sin(x)sin(y)
    1. 0
    2. 1
    3. -1
    4. π
    5. -∞

    Author of lecture Derivative of sin(x)

     Batool Akmal

    Batool Akmal

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