calculus

posted by .

if y=cosx.sinx find dy/dx

  • calculus -

    just use the product rule
    If y = fg, then
    y' = f'g + fg'

    So,

    dy.dx = -sinx.sinx + cosx.cosx
    = cos^2x - sin^2x
    = cos2x

    Or, you could have noticed that
    y = 1/2 sin2x, so
    y' = cos2x

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Math

    Verify the identity . (cscX-cotX)^2=1-cosX/1+cosX _______ sorry i cant help you (cscX-cotX)=1/sinX - cosX/sinX = (1-cosX)/sinX If you square this you have (1-cosX)^2/(sinX)^2 Now use (sinX)^2 = 1 - (cosX)^2 to get (1-cosX)^2 / 1 - …
  2. Pre-Calc

    Trigonometric Identities Prove: (tanx + secx -1)/(tanx - secx + 1)= tanx + secx My work so far: (sinx/cosx + 1/cosx + cosx/cosx)/(sinx/cos x - 1/cosx + cosx/cosx)= tanx + cosx (just working on the left side) ((sinx + 1 - cosx)/cosx)/((sinx …
  3. 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 …
  4. 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 …
  5. 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)
  6. 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!
  7. 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] …
  8. trigonometry

    can i use factoring to simplify this trig identity?
  9. Calculus

    If y=3/(sinx+cosx) , find dy/dx A. 3sinx-3cosx B. 3/(sinx+cosx)^2 C. -3/(sinx+cosx)^2 D. 3(cosx-sinx)/(sinx+cosx)^2 E. 3(sinx-cosx)/(1+2sinxcosx)
  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