Prove that d/dx secx=secxtanx. Find the second derivative of sec x?

Prove that d/dx secx=secxtanx. Find the second derivative of sec x

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  • 9 years ago
    Favorite Answer

    sec(x) = 1/(cos(x))

    let y = 1/(cos(x))

    d/dx 1/cos(x) = by the quotient rule

    y' = [0(cos(x)) - -sin(x)(1)]/cos^2(x)

    = sin(x)/cos^2(x)

    =[sin(x)/(cos(x)][1/cos(x)]

    = tan(x)sec(x)

    To find the second derivative do quotient rule of sin(x)/cos^2(x) or product rule of tan(x)sec(x).

    y'' = sec^2(x)sec(x) + (tan(x)sec(x)tan(x)

    = sec(x)[sec^2(x) + tan^2(x)]

    Source(s): Math Student
  • 5 years ago

    You have several options. For example, if y = tan(x) then arctan(y) = x, so differentiating using the chain rule: (1 / (1+y^2) ) dy/dx = 1 Therefore dy/dx = 1+y^2 = 1 + tan^2 (x) = sec^2 (x). Another way: y = tan(x) = sin(x) / cos(x) = sin(x) ( 1/cos(x) ) = sin(x) sec(x), so use the product rule: dy/dx = sin(x) [sec(x) tan(x)] + sec(x) cos(x) = (sin(x) sec(x)) tan(x) + 1 = (sin(x) (1/cos(x) ) tan(x) + 1 = tan^2 (x) + 1 = sec^2 (x). Finally, you could use a difference quotient: (tan(x+h) - tan(x)) / h =(((tan(h)+tan(x))/(1-tan(h)tan(x))) - tan(x)) / h After simplification and factoring this is tan(h) (1 + tan^2 (x)) / ((1-tan(x) tan(h)) h) Write this as: (tan(h) / h) * (1 / (1-tan(x) tan(h))) * (1 + tan^2 (x)) Now take limit as h-->0. tan(h) / h = (sin(h) / h) (1/cos(h)) --> (1) (1) = 1 as h-->0. 1 / (1-tan(x) tan(h)) --> 1 / (1-tan(x) (0) ) --> 1/1 = 1 as h--> 0 and the third factor (1 + tan^2 (x)) is independent of h, so is unchanged as h-->0. Hence the final answer is (1)(1)((1 + tan^2 (x)) = 1 + tan^2 (x) = sec^2 (x).

  • 9 years ago

    Express as (cos x)^-1

    Then do a simple chain rule differentiation.

    You get sin x/(cos^2)x=sec x tan x

    For second derivative, use product rule.

    Not rocket science.

  • 9 years ago

    One way you can prove that the derivative of sec(x) = sec(x)*tan(x) is by recalling that sec(x) = 1/cos(x) and taking the derivative using quotient rule:

    d/dx[1/cos(x)] = [sin(x)]/cos²(x) = sec(x)*tan(x)

    d/dx[sec(x)*tan(x)] = d²y/dx²[sec(x)] = sec^3(x) + sec(x)*tan^2(x)

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  • 9 years ago

    sec x = 1/cos x

    d/dx 1/cos x

    (0(cos x) - 1(-sin x))/cos^2(x)

    sin x / cos^2(x)

    tan x / cos x

    d/dx (sec x) = tan x sec x

    second derivative of sec x:

    sec x ( sec^2 (x)) + tan x (sec x tan x)

    sec^3 (x) + sec x tan^2 (x)

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