Math Help - How can I show this is equal to 2sec2x?

1. How can I show this is equal to 2sec2x?

I have the equation f(x) = $ln \tan(\frac{\pi}{4} + x)$ and I want to show that the derivative f ' (x) is equal to $2\sec 2x$.

I have gotten an expression for the derivative as $\csc(\frac{\pi}{4} + x) \sec(\frac{\pi}{4} + x)$ but I'm not sure what I can do in order to show this is equal to 2sec2x.

Please could you help me get closer to the answer? Thanks if you know what can be done with this

2. Originally Posted by db5vry
I have the equation f(x) = $ln \tan(\frac{\pi}{4} + x)$ and I want to show that the derivative f ' (x) is equal to $2\sec 2x$.

I have gotten an expression for the derivative as $\csc(\frac{\pi}{4} + x) \sec(\frac{\pi}{4} + x)$ but I'm not sure what I can do in order to show this is equal to 2sec2x.

Please could you help me get closer to the answer? Thanks if you know what can be done with this
$tan({\frac{\pi}{4}}+x) = sec(2x) + tan(2x)$

so you have to differentiate now and simplify a bit:

differentiate $ln(sec(2x) + tan(2x))$ with respect to x.

3. Hello, db5vry!

I have the function: . $f(x) \:=\:\ln \tan\left(\tfrac{\pi}{4} + x\right)$

and I want to show that the derivative is: . $f'(x) \:=\:2\sec 2x$
$\text{Let: }\:\theta \:=\:\tfrac{\pi}{4} + x$

We have: . $f(x) \;=\;\ln\tan\theta$

Then: . $f'(x) \;=\;\frac{\sec^2\!\theta}{\tan\theta} \;=\;\frac{\dfrac{1}{\cos^2\!\theta}}{\dfrac{\sin\ theta}{\cos\theta}} \;=\;\frac{1}{\sin\theta\cos\theta}\;=\;\frac{2}{2 \sin\theta\cos\theta} \;=\;\frac{2}{\sin2\theta}$ .[1]

$\text{Since }\,2\theta \:=\:\tfrac{\pi}{2} + 2x$, the denominator is: . $\sin\left(\tfrac{\pi}{2} + 2x\right)$

. . . . $=\;\sin\left(\tfrac{\pi}{2}\right)\cos(2x) + \cos\left(\tfrac{\pi}{2}\right)\sin(2x) \;=\;1\!\cdot\!\cos(2x) + 0\!\cdot\!\sin(2x) \;=\;\cos(2x)$

Therefore, [1] becomes: . $\frac{2}{\cos2x} \;=\;2\sec2x$