Question

Consider m(x)=Sin^2(x). Investigate the derivative and show that it can be written as 2sinxcosx.

Answer 1

write out sin^2(x) as sin(x)sin(x)

take the derivative using the product rule:
(sin(x)sin(x))' = sin'(x)sin(x) + sin(x)sin'(x)

if you got this far, you know sin'(x) = cos(x)

it is just very simple algebra from there.

Answer 2

try the dummy approach. lets sin^2(x), take sin(x) = w and then we've got sin^2(x) = w^2.... well the derivative of w^2 is 2*w or 2w whatever, okay we should used the chain I mean we've got 2w and 2w= 2(sin x) then multiply it with the derivative of sin(x), which is cos(x). so the problem is solved. derivative of sin^2(x) is 2sin(x)cos(x) or to be more elaborate is also sin^2(x) - cos^2(x).

Answer 3

Hmmm let's see:
Given f(x) = sin^2(x) and you wanna prove that it's equivalent to 2sinxcosx using the derivative:
Well the derivative of sin^2(x) is
Using the product rule: sinx'sinx + sinxsinx'
cosxsinx + sinxcosx
cosx(2sinx)
or just simply 2sinxcosx
That's one way to solve it :D
OR you can think of sinx as u:
Let u = sinx
So rewritten as u^2
The derivative is this is simply : 2uu'
Now using the substitution method:
2uu' = 2sinxsinx'
Since sinx' = cosx
The answer would be 2sinxcos :)