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

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

Question
Differential equations
asked 2020-11-12
\(F(x) = (lnx)^{\cos}x,\) find f'(x) and the equation to the tangent line at (e 1)

Answers (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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