# Calculus

posted by Juliet

differentiate:
y=sin^2(x)- cos^2(x)

I have this:
y'= 2cosx + 2sinx
What do i do next??

1. Reiny

Ahhh, you might recognize that
cos^2 x - sin^2 x = cos 2x

so y=sin^2(x)- cos^2(x)
= - cos 2x

dy/dx = 2sin 2x or 4sinxcosx

If you don't see that identity right away, then

dy/dx = 2(sinx)cosx - 2cosx(-sinx)
= 4sinxcosx or 2sin(2x)

2. Anonymous

Oh! Thank you so much! I completely forget about the double angle identity!!! :)
THANKS!!!

## Similar Questions

1. ### Math, derivatives

Let g(x) = sin (cos x^3) Find g ' (x): The choices are a) -3x^2sinx^3cos(cos x^3) b) -3x^2sinx^3sin(cos x^3) c) -3x^2cosx^3sin(cos x^3) d) 3x^2sin^2(cos x^3) I'm not exactly sure where I should start. Should I begin with d/dx of sin?
2. ### TRIG!

Posted by hayden on Monday, February 23, 2009 at 4:05pm. sin^6 x + cos^6 x=1 - (3/4)sin^2 2x work on one side only! Responses Trig please help! - Reiny, Monday, February 23, 2009 at 4:27pm LS looks like the sum of cubes sin^6 x + cos^6 …
3. ### trigo help!

1. differentiate cos(3/x) 2. differentiate sin(4/x) 3. differentiate 3/{sin(3x+pi)} 4. differentiate pxsin(q/x)where p and q are constants. 5. differentiate xsin(a/x) where a is constant 6. differentiate sec^3(3x^2+1)
4. ### Calc II

For the following question, we need to find the length of the polar curve: r= 2/(1-cosx) from pi/2 </= x </= pi dr/dx= -2sinx/(1-cosx)^2 therefore the length would be expressed as follows: the integral from pi/2 to pi of (sqrt)[(2/(1-cos))^2 …
5. ### RESPONSE to PreCalc Question

h t t p : / / w w w . j i s k h a . c o m / d i s p l a y . c g i ?
6. ### Calc.

Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x) (dy/dx)-(dx/du)= …
7. ### calculus

Differentiate. y= (cos x)^x u= cos x du= -sin x dx ln y = ln(cos x)^x ln y = x ln(cos x) (dy/dx)/(y)= ln(cos x) (dy/dx)= y ln(cos x) = (cos x)^x * (ln cos x) (dx/du)= x(cos x)^(x-1) * (-sin x) = - x sin(x)cos^(x-1)(x) (dy/dx)-(dx/du)= …
8. ### calculus

Evaluate lim->4 sin(2y)/tan(5y) Here is what I have so far. I am not sure the next steps. Can someone help me?
9. ### maths

by equating the coefficients of sin x and cos x , or otherwise, find constants A and B satisfying the identity. A(2sinx + cosx) + B(2cosx - sinx) = sinx + 8cosx I got A = 2, B = 3, which the answers said were correct. However, another …
10. ### Mathematics-Integration

Question: Prove that [integrate {x*sin2x*sin[π/2*cos x]} dx] /(2x-π) } from (0-π) = [ integrate {sin x*cos x*sin[π/2*cos x} dx ] from (0-π). My thoughts on the question: We know that integrate f(x) dx from (0-a) = integrate f(a-x) …

More Similar Questions