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Math Help - Using first principles, find the derivative of y = tanx.

  1. #1
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    Using first principles, find the derivative of y = tanx.

    I am rather terrible at using trigonometric proofs, which is needed in doing this question.

    Question is title of this thread.

    My attempt:
     \frac{tan(x+h) - tanx}{h}
     = \frac{\frac{sin(x+h)}{cos(x+h)} - \frac{sinx}{cosx}}{h}
     = \frac{\frac {sinxcosh + cosxsina}{cosxcosh - sinxsinh} - \frac{sinx}{cosx}}{h}

    I don't know what to do from here. I need to show all my steps to get the answer. I know the answer is  sec^2 x , but don't know how to show it using first principles.

    If you could show me what to do next, I would be very grateful.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Swampscott, MA
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    Quote Originally Posted by Kakariki View Post
    I am rather terrible at using trigonometric proofs, which is needed in doing this question.

    Question is title of this thread.

    My attempt:
     \frac{tan(x+h) - tanx}{h}
     = \frac{\frac{sin(x+h)}{cos(x+h)} - \frac{sinx}{cosx}}{h}
     = \frac{\frac {sinxcosh + cosxsina}{cosxcosh - sinxsinh} - \frac{sinx}{cosx}}{h}

    I don't know what to do from here. I need to show all my steps to get the answer. I know the answer is  sec^2 x , but don't know how to show it using first principles.

    If you could show me what to do next, I would be very grateful.
    Use this identity: \tan(x+h)=\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)}

    So \tan(x+h)-\tan(x)=\frac{\tan(x)+\tan(h)}{1-\tan(x)\tan(h)}-\tan(x)= \frac{\tan(x)+\tan(h)-\tan(x)+\tan^2(x)\tan(h)}{1-\tan(x)\tan(h)}=\frac{\tan(h)(1+\tan^2(x))}{1-\tan(x)\tan(h)}

    Thus, \lim_{h\to0}\frac{\tan(x+h)-\tan(x)}{h}=\lim_{h\to0}\frac{\tan(h)(1+\tan^2(x))  }{h(1-\tan(x)\tan(h))}

    Spoiler:
    \lim_{h\to0}\frac{\tan(h)(1+\tan^2(x))}{h(1-\tan(x)\tan(h))}=\lim_{h\to0}\left(\frac{\sin(h)}{  h}\cdot\frac{\frac{1}{\cos(h)}(1+\tan^2(x))}{1-\tan(x)\tan(h)}\right)= \left(\lim_{h\to 0}\frac{\sin(h)}{h}\right)\cdot\left(\lim_{h\to0}\  frac{\frac{1}{\cos(h)}(1+\tan^2(x))}{1-\tan(x)\tan(h)}\right) =(1)\cdot\left(\frac{1\cdot(1+\tan^2(x))}{1-1\cdot 0}\right)=1+\tan^2(x)

    By a trig identity, 1+\tan^2(x)=\boxed{\sec^2(x)}

    Your professor is cruel.
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