F(x)=(lnx)cosx, find f'(x) and the equation to the tangent line at (e 1)

Key to "F(x)=(lnx)cosx, find f'(x) and the equation to the..."

Cheyanne Leigh

Answered question

2020-11-12

F (x) = (l n x)^{\cos} x,

find f'(x) and the equation to the tangent line at (e 1)

Answer & Explanation

cyhuddwyr9

Skilled2020-11-13Added 90 answers

It is given that
$f (x) = (\ln x)^{\cos x}$
We first find f`(x).
By exponent rule we know that
$a^{b} - e^{b \ln (a)}$
This gives us
$f (x) = e^{\cos (x) \ln (\ln (x))}$
Let us take
$u (x) = \cos (x) \ln (\ln (x))$
Then by chain rule we have
$f ‘ (x) = \frac{d}{d x} (e^{u}) = e^{u} \times \frac{d u}{d x}$
Now note that
$\frac{d u}{d x} = \frac{d}{d x} (\cos (x) \ln (\ln (x))$
$= \frac{d}{d x} (\cos x) \times \ln (\ln (x)) + \frac{1}{\ln x} \times \frac{1}{x} \times \cos x$
$= - \sin (x) \ln (\ln (x)) + \frac{\cos (x)}{x \ln (x)}$
By subtituting value of du/dx in the expression for f`(x) we obtained
$f ‘ (x) = e^{u} \times \frac{d u}{d x} = e^{\cos (x) \ln (\ln (x)) (- \sin (x) \ln (\ln (x)) + \frac{\cos (x)}{x \ln (x)})}$
Now note that
$e^{\cos (x) \ln (\ln (x))} = (e^{\ln (\ln (x))} (\cos x) = (\ln (x))^{\cos x}$
Hence we get
$f ‘ (x) = - (\ln (\ln (x)) \sin (x) - \frac{\cos (x)}{x \ln (x)}) \ln (x)^{\cos (x)}$
Note that $f (e) = 1$ and the given point (e, 1) can be written as (e, f(e)).
Now we have to find equation of tangent line to $f (x) a t x_{0} = e .$
In this case $y_{0} = 1$
Clearly, the slope of the tangent line at the point (e, 1) is given by
$m = f ‘ (e) = \frac{\cos e}{e}$
Thus the equation of the tangent line is
$y - y_{0} m (x - x_{0}) \Rightarrow y - 1 = \frac{\cos (e)}{e} (x - (e))$
Result answer: $f ‘ (x) = - (\ln (\ln (x)) \sin (x) - \frac{\cos (x)}{x \ln (x)} \ln (x)^{\cos (x)})$
$y - 1 \frac{\cos (e)}{e} (x) - (e))$

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