posted by .

Determine the derivative of the function (² = square):

a) y = (x² + x - 1)Sin²x

b) y = (x² + 1)/Sin²x

chain rule:
d/dx sin^2 x = 2(sinx)(cosx) = sin2x

a)
y = (x^2+x-1)sin^2 x
y' = (2x+1)sin^2 x + (x^2+x-1)sin2x

b)
y = (x^2+1)/sin^2 x
y = (2x*sin^2 x - (x^2+1)*2*sinx*cosx]/sin^4 x
= [2x*sinx - (x^2+1)*cosx]/sin^3 x

Thank you Steve

## Similar Questions

1. ### trigonometry

if sin2x=3sin2y, prove that: 2tan(x-y)=tan(x+y) ( here, in sin2x, 2x is an angle., like there's a formula:sin2x=2sinxcosx and sin2y=2sinycosy; ....)
2. ### trig

Verify the trigonometric identity: [(1–sin²x)/sin²x]–[(csc²x–1)/cos²x]= -tan²x I can't figure this out.
3. ### Calculus

Solve identity, (1-sin2x)/cos2x = cos2x/(1+sin2x) I tried starting from the right side, RS: =(cos²x-sin²x)/(1+2sinxcosx) =(cos²x-(1-cos²x))/(1+2sinxcosx) and the right side just goes in circle. May I get a hint to start off?
4. ### calculus

d/dx(e^sinx2x)= a)-cos2xe^sinx2x b)cosxe^sin2x c)2e^sin2x d)2cos2xe^sin2x e)-2cos2xe^sin2x

a) determine the derivative of y = sin2x b) determine the derivative of y = 2SinxCosx c) shoe that derivatives in parts a) and b) are equal d) explain why derivatives in parts a) and be should be equal.

solve for (<_ = less than or equal to / pie = pie sign / -pie = negative pie) 3 sin²x = cos²x ; 0 <_ x < 2pie cos²x - sin²x = sinx ; -pie < x <_ pie