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    Does anyone have any good links to some websites/videos on how to differentiate sin/cos/tan^-1 x?

    I could memorise the end products of each one but I would prefer to know how to derive them myself because I've seen questions come up such as;

    Differentiate Tan^-1(5x) when first stating tan^-1x = tany Or something similar.

    Any help?
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    Start by writing y = cos^-1 x
    This is the same as cos y = x
    Then you can differentiate implicitly with respect to x. Unfortunately I can't go further than that at this stage... see what you get!

    Edit: Or try it with tan/sin in the same form, I just chose cosine because it's the best.
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    (Original post by Shawshank)
    Start by writing y = cos^-1 x
    This is the same as cos y = x
    Then you can differentiate implicitly with respect to x. Unfortunately I can't go further than that at this stage... see what you get!

    Edit: Or try it with tan/sin in the same form, I just chose cosine because it's the best.
    Just adding onto this, there will come a stage in your working where you use the identities \displaystyle \tan x = \frac{\sin x}{\cos x} and \sin^2 x + \cos^2 x = 1 .
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    Always start by reexpressing the equation without the inverse tirg function. for your example it would be like this
     tan(y) = 5x

    Now differentiate, using implicit differentiation on the LHS
     sec^2(y) \frac{dy}{dx} = 5

    Rearrange to get  \frac{dy}{dx} as subject

     \frac{dy}{dx} = \frac{5}{sec^2(y)}

    Final stage is to remove the y terms on the RHS. In this casr substituting in for  sec^2(y) using trig identity  sec^2(y) = 1 + tan^2(y) = 1 + 25x^2 as  tan y = 5x from first part. Final result is therefore

     \frac{dy}{dx} = \frac{5}{1+25x^2}

    Hope that is useful!
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    This is pretty similar, but isn't there a general formula?

    \displaystyle[f^{-1}(x)]'=\frac{1}{f'(f^{-1}(x))}

    In particular, since \tan'(x)=1/\cos^2(x) (using the formula for differentiating a fraction), [\tan^{-1}(x)]'=\cos^2(\tan^{-1}(x)), and then indeed one has to express \cos via \tan to get rid of \tan^{-1}.
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    (Original post by spread_logic_not_hate)
    Always start by reexpressing the equation without the inverse tirg function. for your example it would be like this
     tan(y) = 5x

    Now differentiate, using implicit differentiation on the LHS
     sec^2(y) \frac{dy}{dx} = 5

    Rearrange to get  \frac{dy}{dx} as subject

     \frac{dy}{dx} = \frac{5}{sec^2(y)}

    Final stage is to remove the y terms on the RHS. In this casr substituting in for  sec^2(y) using trig identity  sec^2(y) = 1 + tan^2(y) = 1 + 25x^2 as  tan y = 5x from first part. Final result is therefore

     \frac{dy}{dx} = \frac{5}{1+25x^2}

    Hope that is useful!
    That was incredibly usefull. Thanks, that's what I was looking for somebody to explain it step by step. I'm going to go try it for the other functions.
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    (Original post by RollerBall)
    I'm going to go try it for the other functions.
    it's really quite fun
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    (Original post by RollerBall)
    Does anyone have any good links to some websites/videos on how to differentiate sin/cos/tan^-1 x?

    I could memorise the end products of each one but I would prefer to know how to derive them myself because I've seen questions come up such as;

    Differentiate Tan^-1(5x) when first stating tan^-1x = tany Or something similar.

    Any help?
    Is this Edexcel C3 :confused:
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    (Original post by FZka)
    Is this Edexcel C3 :confused:
    No it's Further Maths.
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    (Original post by Mathematician!)
    No it's Further Maths.
    I see. Thank you.
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    (Original post by spread_logic_not_hate)
    Hope that is useful!
    I tried doing it with  sin^-1(4x)

    I did:
    siny=4x

    cosy \frac{dy}{dx} =4

     \frac{dy}{dx} = \frac {4}{cosy}

    \frac{dy}{dx} = \frac {4}{1-siny}

    \frac{dy}{dx} = \frac {4}{1-4x}

    Now this is wrong. It should be  \frac{dy}{dx}=\frac{1}{\frac {1}{4}\sqrt{1-16x^2}}

    But this is according to my maths teacher who sometimes makes mistakes. But I'm sure I'm wrong. Any ideas?

    EDIT: I realised what I've done wrong.

    I should have done:
    siny=4x

    siny^2=16x^2

    1-cosy^2=16x^2

    1-16x^2=cosy^2

    cosy=\sqrt {1-16x^2}

    Therefore:

    siny=4x

    cosy \frac{dy}{dx} =4

     \frac{dy}{dx} = \frac {4}{cosy}

    \frac{dy}{dx} = \frac {4}{\sqrt{1-16x^2}}
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    (Original post by RollerBall)
    I tried doing it with  sin^-1(4x)



     \frac{dy}{dx} = \frac {4}{cosy}

    \frac{dy}{dx} = \frac {4}{1-siny}
    how did you go from cosy to 1-siny?
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    (Original post by rbnphlp)
    how did you go from cosy to 1-siny?
    Yeah, check my edit. I realised where I went wrong.
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    (Original post by RollerBall)
    Yeah, check my edit. I realised where I went wrong.
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    (Original post by rbnphlp)
    So Trig inverse functions you don't need to square root/square becuase they indirectly get squared when differentiating to sec^2x, but the others I need to square myself right?

    So in theory, cos/sin should always have a square root in it, and tan shouldn't?
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    For y=sin^{-1}(x)

    y=sin^{-1}(x)
    x=sin(y)
    \frac{dx}{dy}=cos(y)
    \frac{dy}{dx}=\frac{1}{cos(y)}
    \frac{dy}{dx}=\frac{1}{\sqrt{1-sin^2(y)}}
    \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}

    For y=cos^{-1}(x)

    y=cos^{-1}(x)
    x=cos(y)
    \frac{dx}{dy}=-sin(y)
    \frac{dy}{dx}=-\frac{1}{sin(y)}
    \frac{dy}{dx}=-\frac{1}{\sqrt{1-cos^2(y)}}
    \frac{dy}{dx}=-\frac{1}{\sqrt{1-x^2}}

    For y=tan^{-1}(x)

    y=tan^{-1}(x)
    x=tan(y)
    \frac{dx}{dy}=1+tan^2(y)
    \frac{dy}{dx}=\frac{1}{1+tan^2(y  )}
    \frac{dy}{dx}=\frac{1}{1+x^2}
 
 
 
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