Expert: Sherry Wallin - 4/12/2009

Question
I am trying to prove that (sin^2(x))' = 2sinxcosx.

I substituted sin^2(x) into the formula (f(x+h)-f(x))/h but im unsure of what to do next. any help would be greatly appreciated

Thanks

Answer
Hi Olivia~
    Thank God you weren't able to prove it because it is not true. The derivative of sin 2x is 2 cos 2x. The sin 2x is 2sinx cosx. In other words the sin of twice the angle is twice the product of the sin of the angle and cos of the angle. Use the chain rule to see that (sin 2x)' is cos 2x* 2= 2 cox 2x.


Now if you want to show the derivative of sin 2x is 2 cos 2x using the definition of the limit this is how you would do it:

let f(x) = sin 2x
[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


Math Prof