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A mathematical problem (trigonometrical identities and differentiation)?

How would you use the derivative of cosx to prove that

d/dx (secx) = secxtanx

I know the derivative of cosx = -sinx, but I keep getting stuck with all my attempts at answering

Any help would be much appreciated!

5 Answers

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  • 1 decade ago
    Favorite Answer

    A'ighty, first write sec(x) = 1/cos(x). Now take the derivative using the Quotient Rule:

    d/dx (1/cos(x)) = (cos(x)*d/dx(1) - 1*d/dx(cos(x))) / cos^2(x)

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

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

    = (1/cos(x)) * (sin(x) / cos(x))

    = sec(x)*tan(x).

    And there you go. ^_^

  • 1 decade ago

    Just observe that sec(x) = 1/cos(x) = cos(x)^(-1), whenever cos (x) <> 0. Now, by the chain rule, it follows that d/dx(secx(x) = (-1) * cos(x)^(-2) * d/dx(cos(x) = (-1) * (1/cos (x)) * (1/cos(x)) * (-sin(x) = sec (x) * sin(x)/cos(x). Since sin(x)/cos(x) = tan(x), we gfinally have that d/dx(sec(x)) = sec(x) * tan(x)

    Source(s): some knowledge about Math
  • Anonymous
    1 decade ago

    d(u/v)=(vdu-udv)/v^2...(1)

    (quotient rule)

    secx=1/cosx

    let u=1,v=cosx

    substitute into (1)

    d(1/cosx)=(cosx*0-1*(-sinx))

    /cosx*cosx

    =(0 +sinx)/

    cosx*cosx

    =(sinx/cosx)*1/cosx

    =tanx*secx

    as required

    i hope that this helps

  • 5 years ago

    (a million - cosA)/sinA + sinA/(a million - cosA) = 2cscA hint: Please determine you employ areas and parentheses to make your question sparkling. :) Yahoo!solutions truncates the question regardless of if this is basically too long and would not contain areas. Take the LHS. (a million - cosA)/sinA + sinA/(a million - cosA) = the effortless denominator is sinA(a million - cosA), so multiply as mandatory. [(a million - cosA)/sinA * (a million - cosA)/(a million - cosA)] + [sinA/(a million - cosA) * (sinA/sinA)] = [(a million - cosA)(a million - cosA)/sinA(a million - cosA)] + [sin²A/(a million - cosA)sinA] = Simplify. [(a million - 2cosA + cos²A)/sinA(a million - cosA)] + [sin²A/(a million - cosA)sinA] = (a million - 2cosA + cos²A+ sin²A) / (a million - cosA)sinA = remember that cos²A+ sin²A = a million (a million - 2cosA + a million) / (a million - cosA)sinA = (2 - 2cosA) / (a million - cosA)sinA = element. 2(a million - cosA) / (a million - cosA)sinA = The (a million - cosA) cancel. 2 / sinA = 2 * a million/sinA = remember that cscA = a million/sinA 2 * cscA = 2cscA = RHS desire this helped!

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  • 1 decade ago

    d/dx (sec x) =

    d/dx (1 / cos x)

    Quotient rule

    (f/g)' = (f'g - fg') / (g²)

    f(x) = 1; g(x) = cos x

    f'(x) = 0; g'(x) = -sin x

    d/dx (1 / cos x) = ( 0 - (-sin x) ) / cos² x

    = sin x / cos² x

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

    = sec x tan x

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