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Math Help - Avg. rate of change of f(x)=tan(x)...

  1. #1
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    Avg. rate of change of f(x)=tan(x)...

    in the interval [x,x+h]

    f(x)=tan(x)

    Avg. rate of change = (f(x+h)-f(x))/(x+h-x)\

    = (tan(x+h) - tan(x))/h

    I need this to equal (tanh/h)*((sec^2(x))/(1-tanxtanh))

    I'm thinking tan addition forumula...

    [((tanx+tanh)/(1-tanxtanh)) -tanx]/h

    I tried common denominator on the top, factoring out a tanh, but I'm going in circles. Please help.
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  2. #2
    Senior Member eumyang's Avatar
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    \begin{aligned}<br />
\dfrac{\tan (x+h) - \tan (x)}{h} &= \dfrac{1}{h}(\tan (x+h) - \tan (x)) \\<br />
&= \dfrac{1}{h}\left( \dfrac{\tan\,x + \tan\,h}{1 - \tan\,x\,\tan\,h} - \tan (x) \right) \\<br />
&= \dfrac{1}{h}\left( \dfrac{\tan\,x + \tan\,h}{1 - \tan\,x\,\tan\,h} -  \dfrac{\tan\,x(1 - \tan\,x\,\tan\,h)}{1 - \tan\,x\,\tan\,h} \right) \\<br />
&= \dfrac{1}{h}\left( \dfrac{\tan\,x + \tan\,h - \tan\,x + \tan^2\,x\,\tan\,h}{1 - \tan\,x\,\tan\,h} \right) \\<br />
&= \dfrac{1}{h}\left( \dfrac{\tan\,h +  \tan^2\,x\,\tan\,h}{1 - \tan\,x\,\tan\,h} \right) \\<br />
&= \dfrac{1}{h}\left( \dfrac{\tan\,h(1 +  \tan^2\,x)}{1 - \tan\,x\,\tan\,h} \right) \\<br />
 &= \dfrac{\tan\, h}{h} \left( \dfrac{1 +  \tan^2\,x}{1 - \tan\,x\,\tan\,h}  \right) \\<br />
 &= \dfrac{\tan\, h}{h} \left( \dfrac{\sec^2\,x}{1 - \tan\,x\,\tan\,h} \right) \\<br />
 \end{aligned}
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