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Math Help - Proving 2 derivatives equal to each other.

  1. #1
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    Proving 2 derivatives equal to each other.

    f(x)=sec2x and g(x)=tan2x. Show that f'(x)=g'(x)

    Well I didn't know how to do secant or tangent very well.
    So I converted it into sine and cosine.
    sec2x = 1/cos2x
    Then I took the derivate of cosine
    f'(x)=1/-sin2x

    Then somehow I think I'm supposed to get g(x) into that as well?

    Can someone explain it to me so I could understand it better.


    P.S. Also does anyone know of a good website for trigonometri functions that convert sine, cosines, tagent, etc.
    Like for example, like sinx/x = 1, or 1-cosx/cox=0, then there are a bunch of other rules. Anyone know the website of that? I used to have them in my notes, but then I lost the paper
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  2. #2
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    Re: Proving 2 derivatives equal to each other.

    Quote Originally Posted by Chaim View Post
    f(x)=sec2x and g(x)=tan2x. Show that f'(x)=g'(x)

    Well I didn't know how to do secant or tangent very well.
    So I converted it into sine and cosine.
    sec2x = 1/cos2x
    Then I took the derivate of cosine
    f'(x)=1/-sin2x

    Then somehow I think I'm supposed to get g(x) into that as well?

    Can someone explain it to me so I could understand it better.


    P.S. Also does anyone know of a good website for trigonometri functions that convert sine, cosines, tagent, etc.
    Like for example, like sinx/x = 1, or 1-cosx/cox=0, then there are a bunch of other rules. Anyone know the website of that? I used to have them in my notes, but then I lost the paper
    I don't know of a web site of hand but bu the pythagorean identity we have that

    \tan^2(x)+1=\sec^2(x)

    So note that the two functions only differ by a constant.

    Also your derivative is incorrect. You would need to use the quotient rule to take the derivative.

    f(x)=\frac{1}{\cos^2(x)} \implies f'(x)=\frac{0\cdot \cos^2(x)-1\cdot 1\cdot \sin(x)\cos(x)}{\cos^4(x)}=-\sec^2(x)\tan(x)
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  3. #3
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    Re: Proving 2 derivatives equal to each other.

    you are expected to know the derivative of \sec{x} and \tan{x} ... look them up and/or derive them.


    f(x) = \sec^2{x} = (\sec{x})^2

    chain rule ...

    f'(x) = 2(\sec{x}) \cdot \sec{x}\tan{x}



    g(x) = \tan^2{x} = (\tan{x})^2

    chain rule again ...

    g'(x) = 2(\tan{x}) \cdot \sec^2{x}
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    Re: Proving 2 derivatives equal to each other.

    Hello, Chaim!

    Your differentiation is incorrect . . . way off!


    f(x)\:=\:\sec^2\!x\,\text{ and }\,g(x)\:=\:\tan^2\!x.\;\;\text{ Show that }f'(x)\:=\:g'(x)


    We have: . f(x) \:= \: (\sec x)^2 \:=\:\frac{1}{(\cos x)^2} \:=\: (\cos x)^{-2}

    Then: . f'(x) \:=\:\text{-}2(\cos x)^{-3}(\text{-}\sin x) \:=\:\boxed{\frac{2\sin x}{\cos^3x}}


    We have: . g(x) \:=\: (\tan x)^2 \:=\: \frac{(\sin x)^2}{(\cos x)^2}

    Then: . g'(x) \;=\;\frac{\cos^2\!x\cdot 2\sin x\cos x - \sin^2\!x\cdot 2\cos x(\text{-}\sin x)}{\cos^4\!x}

    n . . . . g'(x) \;=\;\frac{2\sin x\cos^3\!x + 2\sin^3\!x\cos x}{\cos^4\!x}

    n . . . . g'(x) \;=\;\frac{2\sin x\cos x\overbrace{(\cos^2\!x + \sin^2\!x)}^\text{This is 1}}{\cos^4\!x}

    n . . . . g'(x) \;=\; \frac{2\sin x\cos x}{\cos^4\!x} \;=\; \boxed{\frac{2\sin x}{\cos^3\!x}}
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