# F(x) = (lnx)^{cos}x, find f'(x) and the equation to the tangent line at (e 1)

$F\left(x\right)=\left(lnx{\right)}^{\mathrm{cos}}x,$ find f'(x) and the equation to the tangent line at (e 1)
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cyhuddwyr9

It is given that
$f\left(x\right)=\left(\mathrm{ln}x{\right)}^{\mathrm{cos}x}$
We first find f(x).
By exponent rule we know that

This gives us

Let us take
$u\left(x\right)=\mathrm{cos}\left(x\right)\mathrm{ln}\left(\mathrm{ln}\left(x\right)\right)$
Then by chain rule we have
$f‘\left(x\right)=\frac{d}{dx}\left({e}^{u}\right)={e}^{u}×\frac{du}{dx}$
Now note that
$\frac{du}{dx}=\frac{d}{dx}\left(\mathrm{cos}\left(x\right)\mathrm{ln}\left(\mathrm{ln}\left(x\right)\right)$

By subtituting value of du/dx in the expression for f(x) we obtained

Now note that
${e}^{\mathrm{cos}\left(x\right)\mathrm{ln}\left(\mathrm{ln}\left(x\right)\right)}=\left({e}^{\mathrm{ln}\left(\mathrm{ln}\left(x\right)\right)}\left(\mathrm{cos}x\right)=\left(\mathrm{ln}\left(x\right){\right)}^{\mathrm{cos}x}$
Hence we get
$f‘\left(x\right)=-\left(\mathrm{ln}\left(\mathrm{ln}\left(x\right)\right)\mathrm{sin}\left(x\right)-\frac{\mathrm{cos}\left(x\right)}{x\mathrm{ln}\left(x\right)}\right)\mathrm{ln}\left(x{\right)}^{\mathrm{cos}\left(x\right)}$
Note that $f\left(e\right)=1$ and the given point (e, 1) can be written as (e, f(e)).
Now we have to find equation of tangent line to
In this case ${y}_{0}=1$
Clearly, the slope of the tangent line at the point (e, 1) is given by
$m=f‘\left(e\right)=\frac{\mathrm{cos}e}{e}$
Thus the equation of the tangent line is
$y-{y}_{0}m\left(x-{x}_{0}\right)⇒y-1=\frac{\mathrm{cos}\left(e\right)}{e}\left(x-\left(e\right)\right)$
Result answer: $f‘\left(x\right)=-\left(\mathrm{ln}\left(\mathrm{ln}\left(x\right)\right)\mathrm{sin}\left(x\right)-\frac{\mathrm{cos}\left(x\right)}{x\mathrm{ln}\left(x\right)}\mathrm{ln}\left(x{\right)}^{\mathrm{cos}\left(x\right)}\right)$
$y-1\frac{\mathrm{cos}\left(e\right)}{e}\left(x\right)-\left(e\right)\right)$