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

1. ## 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))

[((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.

2. \begin{aligned}
\dfrac{\tan (x+h) - \tan (x)}{h} &= \dfrac{1}{h}(\tan (x+h) - \tan (x)) \\
&= \dfrac{1}{h}\left( \dfrac{\tan\,x + \tan\,h}{1 - \tan\,x\,\tan\,h} - \tan (x) \right) \\
&= \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) \\
&= \dfrac{1}{h}\left( \dfrac{\tan\,x + \tan\,h - \tan\,x + \tan^2\,x\,\tan\,h}{1 - \tan\,x\,\tan\,h} \right) \\
&= \dfrac{1}{h}\left( \dfrac{\tan\,h + \tan^2\,x\,\tan\,h}{1 - \tan\,x\,\tan\,h} \right) \\
&= \dfrac{1}{h}\left( \dfrac{\tan\,h(1 + \tan^2\,x)}{1 - \tan\,x\,\tan\,h} \right) \\
&= \dfrac{\tan\, h}{h} \left( \dfrac{1 + \tan^2\,x}{1 - \tan\,x\,\tan\,h} \right) \\
&= \dfrac{\tan\, h}{h} \left( \dfrac{\sec^2\,x}{1 - \tan\,x\,\tan\,h} \right) \\
\end{aligned}