## anonymous 3 years ago The question for Session4 requires you to find the derivative of sin(2x). I tried using various sin/cos properties but nothing was able to cancel out the deltaX in the denominator. I tried things like sin(x+y) = sin(x)cos(y)+cos(x)sin(y), and cos(x+y) = cos(x)cos(y)-sin(x)sin(y), and sin(x)^2 = 1 - cos(x)^2. Does anyone know the property I should examine in order to help pull out the deltaX?

1. anonymous

Isn't the derivative of Sin(2x) just 2Cos(2x) ?

2. anonymous

Yeah, looking at a derivatives table I see that sin(ax) is acos(ax) but I was curious as to how to get to that conclusion. I wanted to find df/dx of sin(2x) the same way Newton did :p. I suppose I need to look back into more trig properties. Do you know of any that are particularly useful in this case. I have a feeling cos^2 + sin^2 = 1 is at play here, but I can't put my finger on it. What do you think?

3. anonymous

Oh you would just find the derivative using first principle, let f(x) = sin 2x so f(x+h) = sin(2x + h) [f(x+h) - f(x)]/h = [sin(2x + h) - sin(2x)]/h now use sin(a+b) = sin a cos b + cos a sin b so sin(2x+h) = sin 2x cos h + cos 2x sin h lim(h->0)[sin 2x cos 2h + cos 2x sin 2h - sin 2x]/h = lim(h->0) [sin 2x cos 2h - sin 2x]/h + lim(h->0) [cos 2x sin 2h]/h = sin 2x *lim(h->0) [cos 2h - 1]/h + cos 2x * lim(h->0) [sin 2h]/h = sin 2x * 0 + cos 2x * lim(h->0)[2 sin h cos h]/h = cos 2x * 2*lim(h->0)[sin h]/h * lim(h->0) cos h = 2cos 2x (1)(1) = 2cos 2x

4. anonymous

When working it out as a limit of a difference quotient, I think it's probably similar to the problem of finding the derivative of sin(x), which is dealt with in Session 7, though that still leaves a couple of tricky things for Session 8. When you've also learnt about the chain rule, you'll be able to use that in order to reduce the problem to the problem of differentiating sin(x).

5. anonymous

Ahh, great feedback! I can see how d/dx of cosx at 0 is zero geometrically speaking just by looking at a cos curve but I'm lost at how lim(h->0) [cosh - 1]/h goes to 0. Is that simply because the constant (-1) goes away as a rule? I can also see how lim(h->0)sinh/h at 0 might be 1 by looking at the sine curve but I feel the need to prove it somehow. Hmm... what do you think? I recon there is a proof that the slope of sin at 0 is 1 but maybe my skills will be lacking until session 8.

6. anonymous

What's taught in session 8?

7. anonymous

SomeBloke mentioned the chain rule in Session 8 being used to differentiate sinx.

8. anonymous

Oops! I could have been clearer. The chain rule is introduced in Session 11. It lets you break differentiations down into a chain of smaller, easier, more manageable steps. Session 7 is the session in which sin(x) and cos(x) are differentiated, but the proofs rely on stuff done in Session 8. Session 8 shows the following:-$\lim_{x \rightarrow 0}\frac{ \sin x }{ x }=1$$\lim_{x \rightarrow 0}\frac{ 1-\cos x }{ x }=0.$The proofs make use of geometry. You might also like to look up the squeeze theorem on Wikipedia. That article gives those two limits as its second example of applying the squeeze theorem, though it doesn't go into detail for them.

9. anonymous

Let z=2x by the chain rule d(sin 2x)/dx = (d(sin z)/dz)*(dz/dx) = (cos z)*(d(2x)/dx)=(cos 2x)*2=2cos(2x)

10. anonymous

Derivative of [f(x)=\sin(2x)\] Derivative of $\sin(x)$ is $\cos(x)$. By the chain rule you, in simple terms, do the derivative of the outside multiplied by the derivative of the inside. The derivative of the outside is cos(2x), while the derivative of the inside is 2, multiply them together and.... the final answer is $f'(x)=2\cos(2x)$.

11. anonymous

sin2x=2sinxcosx...now integrate by parts