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: \(f`(x) = -(\ln(\ln(x)) \sin (x) - \frac{\cos(x)}{x \ln (x)}\ln (x)^{\cos(x)})\)

\(y - 1 \frac{\cos(e)}{e} (x)-(e))\)

\(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: \(f`(x) = -(\ln(\ln(x)) \sin (x) - \frac{\cos(x)}{x \ln (x)}\ln (x)^{\cos(x)})\)

\(y - 1 \frac{\cos(e)}{e} (x)-(e))\)