Calculus

posted by .

Find the derivative of
f(x) =cos^-1(x^2)

Ok so I'm having a little confusion with the chain rule and I'm not sure if my answer is right...

will the answer be... 2x(-cos^-2(x^2)(-sin(x^2)?

  • Calculus -

    recall that the derivative of cos^-1 (x) or arccos (x) is
    -1/[sqrt(1-x^2)] * dx
    thus the derivative of cos^-1 (x^2)
    -(2x)/[sqrt(1 - x^4)]

    note that the 2x is the dx that came from the derivatice of the term inside the cos^-1, which is x^2.

    hope this helps~ :)

  • Calculus -

    In google type:
    calc101

    When you see list of results click on first link.

    When page be open click option derivatives.

    When that page be open in rectangle type your function and click option DO IT.

    You will see solution step by step.

  • Calculus -

    If cos^-1(x^2) mean 1/cos(x^2)

    type cos[x^2]^(-1)


    If cos^-1(x^2) mean arccos(x^2)

    type arccos[x^2]

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Mathematics - Trigonometric Identities

    Let y represent theta Prove: 1 + 1/tan^2y = 1/sin^2y My Answer: LS: = 1 + 1/tan^2y = (sin^2y + cos^2y) + 1 /(sin^2y/cos^2y) = (sin^2y + cos^2y) + 1 x (cos^2y/sin^2y) = (sin^2y + cos^2y) + (sin^2y + cos^2y) (cos^2y/sin^2y) = (sin^2y …
  2. Math, derivatives

    f(x) = x² + 2Cos²x, find f ' (x) a) 2(x+cos x) b) x - sin x c) 2x + sin x d)2(x - sin2x) I got neither of these answers, since the 2nd part should be chain rule, right?
  3. 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
  4. Math

    Solve this equation algebraically: (1-sin x)/cos x = cos x/(1+sin x) --- I know the answer is an identity, and when graphed, it looks like cot x. I just don't know how to get there. I tried multiplying each side by its conjugate, but …
  5. calculus

    Find the points on the curve y= (cos x)/(2 + sin x) at which the tangent is horizontal. I am not sure, but would I find the derivative first: y'= [(2 + sin x)(-sin x) - (cos x)(cos x)]/(2 + sin x)^2 But then I don't know what to do …
  6. Calculus 12th grade (double check my work please)

    1.)Find dy/dx when y= Ln (sinh 2x) my answer >> 2coth 2x. 2.)Find dy/dx when sinh 3y=cos 2x A.-2 sin 2x B.-2 sin 2x / sinh 3y C.-2/3tan (2x/3y) D.-2sin2x / 3 cosh 3yz...>> my answer. 2).Find the derivative of y=cos(x^2) …
  7. Calculus

    Use implicit differentiation to find dy/dx. e^4x = sin(x+2y). This is a practice problem. It says the correct answer is 4e^x/(2sin(x+2y)) but I keep getting 4e^(4x)/(2cos(x+2y)). I thought the derivative of e^(4x) would be 4e^(4x), …
  8. Cos-Derivative

    y= (cos^3 x) (cos 3x) I got -3 sin(3x) cos^3x - 3 sin(x) cos (3x) cos^2 (x) using the product rule Is this right?
  9. Calculus

    I have y = sin^2(3^(x)) which I rewrite as y = (sin(3^(x)))^2 I got the derivative using the chain rule as the following y' = 3(sin(3^x))^2 (cos(3x)) (3^(x)(ln(3)) Can anyone confirm whether this is right/wrong?
  10. Calculus

    Find f'(x) if f(x)=sin^3(4x) A. 4cos^3(4x) B. 3sin^2(4x)cos(4x) C. cos^3(4x) D. 12sin^2(4x)cos(4x) E. None of these I got D using the chain rule?

More Similar Questions