finding derivative of g(x) = xtanx
By Sophia Hammond
The question reads if $g(x) = x\tan x$, then the value of $g'\left(\frac\pi4\right)$ is:
a) $1+ \frac\pi4$
b) $\frac\pi2-2$
c) $1-\frac\pi4$
d) $1+\frac\pi2$
I can never get to a situation where I have all trig functions out of the derivative as is shown in any of the answers
$\endgroup$ 14 Answers
$\begingroup$Note; \begin{equation} g'(x)=\tan x + x \sec^{2}x \end{equation} At $\pi/4$ \begin{eqnarray} g'(\pi/4) &=& \tan \frac{\pi}{4}+\frac{\pi}{4} \sec^{2} \frac{\pi}{4}\\ &=& 1+ \frac{\pi}{4} \left(\frac{2}{\sqrt{2}} \right)^{2}\\ &=& 1+\frac{\pi}{2} \end{eqnarray}
$\endgroup$ $\begingroup$HINT:
First, calculate $g'(x)$ for a general value of $x$. For that, write down $g$ as a product of two functions you know how to differentiate:
$$g(x)=f(x)\cdot h(x)$$
This part should be easy as the choice of $f$ and $h$ is quite apparent.
Then use the product rule:
$$g'(x) = f'(x)h(x) + f(x)h'(x)$$
Then, once you have $g'(x)$, just plug in $x=\frac\pi4$.
Try to use my hint and see how far you get. If you are still stuck, then edit your question with your progress (or comment on this question) and I (we) will attempt to help you further.
$\endgroup$ 3 $\begingroup$$$g'(x)=\tan{x}+x(1+\tan^2{x})$$
so $g'(\frac{\pi}{4})=1+\frac{\pi}{2}$
$\endgroup$ $\begingroup$$$g(x)=x\tan x$$ $$g'(x)=\tan x+\frac{x}{\sin^2x}$$ $$g'(\pi/4)=\tan(\pi/4)+\frac{\pi/4}{\sin^2(\pi/4)}=1+\frac{\pi}{4(\frac{\sqrt 2}{2})^2}=1+\pi/2$$
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