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

Cheyanne Leigh 2020-11-12 Answered
F(x)=(lnx)cosx, find f'(x) and the equation to the tangent line at (e 1)
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Expert Answer

cyhuddwyr9
Answered 2020-11-13 Author has 90 answers

It is given that
f(x)=(lnx)cosx
We first find f`(x).
By exponent rule we know that
abeb ln(a)
This gives us
f(x)=ecos(x) ln(ln(x))
Let us take
u(x)=cos(x)ln(ln(x))
Then by chain rule we have
f(x)=ddx(eu)=eu×dudx
Now note that
dudx=ddx(cos(x)ln(ln(x))

=ddx(cosx)×ln(ln(x))+1ln x×1x×cosx
=sin(x)ln(ln(x))+cos(x)x ln(x)
By subtituting value of du/dx in the expression for f`(x) we obtained
f(x)=eu×dudx=ecos(x)ln(ln(x))(sin(x)ln(ln(x))+cos(x)x ln (x))
Now note that
ecos(x)ln(ln(x))=(eln(ln(x))(cosx)=(ln(x))cosx
Hence we get
f(x)=(ln(ln(x))sin(x)cos(x)xln(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 x0=e.
In this case y0=1
Clearly, the slope of the tangent line at the point (e, 1) is given by
m=f(e)=cosee
Thus the equation of the tangent line is
yy0m(xx0)y1=cos(e)e(x(e))
Result answer: f(x)=(ln(ln(x))sin(x)cos(x)xln(x)ln(x)cos(x))
y1cos(e)e(x)(e))

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