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

Question
Differential equations
$$F(x) = (lnx)^{\cos}x,$$ find f'(x) and the equation to the tangent line at (e 1)

2020-11-13
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}{dx} (e^{u}) = e^{u} \times \frac{du}{dx}$$
Now note that
\frac{du}{dx} = \frac{d}{dx} (\cos (x) \ln (\ln(x)))\)

$$= \frac{d}{dx} (\cos x) \times \ln(\ln(x)) + \frac{1}{\ln\ x} \times \frac{1}{x} \times \cos x$$
$$= - \sin (x) \ln (\ln(x)) + \frac{\cos (x)}{x\ \ln (x)}$$
By subtituting value of du/dx in the expression for f(x) we obtained
$$f(x) = e^{u} \times \frac{du}{dx} = e^{\cos(x)\ln(\ln(x)) (-\sin (x) \ln(\ln(x)) + \frac{\cos(x)}{x\ \ln\ (x)})}$$
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 $$f(x)\ at\ x_{0} = e.$$
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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