Math

posted by .

Find the derivative of
y=sin(x)*(sinx+cosx)

  • Math -

    just use the product rule

    If y = uv then
    y' = u'v + uv'
    where u=sinx and v=sinx+cosx

    y' = cosx(sinx+cosx) + sinx(cosx-sinx)
    = sin2x + cos2x

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Trigonometry.

    ( tanx/1-cotx )+ (cotx/1-tanx)= (1+secxcscx) Good one! Generally these are done by changing everything to sines and cosines, unless you see some obvious identities. Also generally, it is best to start with the more complicated side …
  2. Trig........

    I need to prove that the following is true. Thanks (cosx / 1-sinx ) = ( 1+sinx / cosx ) I recall this question causing all kinds of problems when I was still teaching. it requires a little "trick" L.S. =cosx/(1-sinx) multiply top and …
  3. Mathematics - Trigonometric Identities

    Prove: (tanx)(sinx) / (tanx) + (sinx) = (tanx) - (sinx) / (tanx)(sinx) What I have so far: L.S. = (sinx / cosx) sinx / (sinx / cosx) + sinx = (sin^2x / cosx) / (sinx + (sinx) (cosx) / cosx) = (sin^2x / cosx) / (cosx / sinx + sinxcosx)
  4. calculus

    Could someone check my reasoning? thanx Find the derivative of the function. sin(sin[sinx]) I need to use the chain rule to solve. So I take the derivative sin(sin[sinx) first. Then multiply that by the inside which is the derivative
  5. Math - Pre- Clac

    Prove that each of these equations is an identity. A) (1 + sinx + cos x)/(1 + sinx + cosx)=(1 + cosx)/sinx B) (1 + sinx + cosx)/(1 - sinx + cosx)= (1 + sin x)/cosx Please and thankyou!
  6. Trigonometry Check

    Simplify #3: [cosx-sin(90-x)sinx]/[cosx-cos(180-x)tanx] = [cosx-(sin90cosx-cos90sinx)sinx]/[cosx-(cos180cosx+sinx180sinx)tanx] = [cosx-((1)cosx-(0)sinx)sinx]/[cosx-((-1)cosx+(0)sinx)tanx] = [cosx-cosxsinx]/[cosx+cosxtanx] = [cosx(1-sinx]/[cosx(1+tanx] …
  7. Math (trigonometric identities)

    I was given 21 questions for homework and I can't get the last few no matter how hard and how many times I try. 17. Sinx-1/sinx+1 = -cos^2x/(sinx+1)^2 18. Sin^4x + 2sin^2xcos^2x + cos^4x = 1 19. 4/cos^2x - 5 = 4tan^2x - 1 20. Cosx …
  8. Trigonometry

    Prove the following trigonometric identities. please give a detailed answer because I don't understand this at all. a. sin(x)tan(x)=cos(x)/cot^2 (x) b. (1+tanx)^2=sec^2 (x)+2tan(x) c. 1/sin(x) + 1/cos(x) = (cosx+sinx)(secx)(cscx) d. …
  9. Precalculus/Trig

    I can't seem to prove these trig identities and would really appreciate help: 1. cosx + 1/sin^3x = cscx/1 - cosx I changed the 1: cosx/sin^3x + sin^3x/sin^3x = cscx/1-cosx Simplified: cosx + sin^3x/sin^3x = cscx/1-cosx I don't know …
  10. Calculus 2 Trigonometric Substitution

    I'm working this problem: ∫ [1-tan^2 (x)] / [sec^2 (x)] dx ∫(1/secx)-[(sin^2x/cos^2x)/(1/cosx) ∫cosx-sinx(sinx/cosx) ∫cosx-∫sin^2(x)/cosx sinx-∫(1-cos^2(x))/cosx sinx-∫(1/cosx)-cosx sinx-∫secx-∫cosx …

More Similar Questions