posted by .

Can someone help me find the derivatives of these questions?

For these can you only need to tell me the answer so i can see if i have the same thing.

y = sinx^2

y = xcosx

I need help with these 6 please.

y = xcos(1/x)

y = (1+sinx)/(1-sin^(2)x)

y = sin^2x/cosx

y = sqrt(sin(sqrtx))

y = 3sqrt(xcosx)

y = 1/(sin(x-sinx)

  • Calculus -

    y' = -4(-sin(-4x)
    = 4sin(-4x)

    y = sinx^2 or (sinx)^2

    y' = 2sinxcosx or sin 2x

    y = xcosx
    y' = x(-sinx) + (1)cosx
    = cosx - xsinx

  • Calculus -

    for y = xcos(1/x)
    I would use the product rule

    y' = x(-sin(1/x))(-1/x^2) + (1)cos(1/x)

    for y = (1+sinx)/(1-sin^(2)x) I would reduce it to

    y = (1+sinx)/((1+sinx)(1-sinx))
    = 1/(1-sinx)
    = (1-sinx)^-1

    now y' = -1(1-sinx)-2?sup>(-cosx)

    why don't you show me some of the steps you have sofar for the others?

  • Calculus -

    last line should have come out as ...

    y = -1(1-sinx)-2(-cosx)

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. calculus

    find the following derivatives: f(x) = (2x+5)/(x^2-3) f'(x)= (2)*(x^2 -3)-(2x+5)*2x/(x^2-3)^2 is that correct?
  2. 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
  3. maths - trigonometry

    I've asked about this same question before, and someone gave me the way to finish, which I understand to some extent. I need help figuring out what they did in the second step though. How they got to the third step from the second. …
  4. Trig Help

    Prove the following: [1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)] =[sinx+sin^2x]/[sinx+1] =[sinx+(1-cos^2x)]/[sinx+1] =?
  5. 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 …
  6. Check a few more CALC questions, please?

    I feel pretty good on my answers. Are there any I got wrong?
  7. Calculus

    Find the derivatives of the following 18. y=〖cos〗^4 x^4 ANSWER: tanx2 19. y= sinx/(1+ 〖cos〗^2 x) ANSWER: 3sinx 20. y=sinx(sinx+cosx) ANSWER: sinx Can you check my answers?
  8. Math Help

    Hello! Can someone please check and see if I did this right?
  9. Calculus

    Just got through with derivatives, power rule, and chain rule. I am supposed to find the first and second derivatives of the function. y = 4x/√x+1 and y = cos^2(x) y' = [(x+1)^(1/2)(4)]-[(4x)(1/2(x+1)^(-1/2))(1)] y' = [4(x+1)^(1/2)-4x]/[2(x+1)^(1/2)(x+1)] …
  10. Calculus

    Integrate 1/sinx dx using the identity sinx=2(sin(x/2)cos(x/2)). I rewrote the integral to 1/2 ∫ 1/(sin(x/2)cos(x/2))dx, but I don't know how to continue. Thanks for the help. Calculus - Steve, Tuesday, January 12, 2016 at 12:45am …

More Similar Questions