derivative

That used me 75 minutes and I didn't use any software......

Too many steps to include here, so here's the result given by Mathcad:

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That's really a complicated one......

I will give it a go.

\(f(x)=\dfrac{3^x\ln x\sqrt{(\ln x)(x^{\sqrt{3}-1})}}{2^x\sin x \cot x}=(\dfrac{3}{2})^x(\dfrac{\ln x\sqrt{(\ln x)(x^{\sqrt3-1})}}{\cos x})\)

\(=(\dfrac{3}{2})^x(\dfrac{\ln x}{\cos x})(\sqrt{(\ln x)(x^{\sqrt3-1})})\)

There we seperate it into 3 parts!!

Part 1 = (3/2)^x

Part 2 = (ln x)/(cos x)

Part 3 = sqrt((ln x)(x^(sqrt 3 - 1)))

Next we will find the derivative of (3/2)^x using implicit differentiation......

\(y=(\dfrac{3}{2})^x\\ \ln y = x\ln \dfrac{3}{2}\\ \dfrac{1}{y}\dfrac{dy}{dx}=\ln\dfrac{3}{2}\\ \dfrac{dy}{dx}=y\ln\dfrac{3}{2}=(\dfrac{3}{2})^x(\ln\dfrac{3}{2})\)

Then we find the derivative of ln x/ cos x......

\((\dfrac{\ln x}{\cos x})'\\ = \dfrac{(\frac{1}{x})(\cos x)-(-\sin x)(\ln x)}{\cos^2 x}\\ =\dfrac{\frac{\cos x}{x}+\sin x \ln x}{\cos^2x}\\ =\dfrac{1}{x\cos x}+\dfrac{\tan x \ln x}{\cos x}\\ =\dfrac{1+x\tan x \ln x}{x\cos x}\)

Then we came to the most complicated part: derivative of sqrt((ln x)(x^(sqrt 3 - 1)))

\((\sqrt{(\ln x)(x^{\sqrt3-1})})'\\ y=\sqrt{u}\\ u = (\ln x)(x^{\sqrt3-1})\\ y' = \dfrac{({x^{\sqrt3-2}})+((\sqrt3-1)(\ln x)(x^{\sqrt3-2}))}{2\sqrt{\ln x(x^{\sqrt3 -1})}}\)

Next we apply the product rule to Part 1 and Part 2 and call it Part 1,2

(Part 1 times Part 2)' 

= Part 1 ' Part 2 + Part 2 ' Part 1

= \(\left(\left(\dfrac{3}{2}\right)^x\ln \dfrac{3}{2}\right)\left(\dfrac{\ln x}{\cos x}\right)+\left(\dfrac{1+x\tan x \ln x}{x\cos x}\right)\left(\dfrac{3}{2}\right)^x\\ \)

= \(\left(\dfrac{3^x\ln x\ln 3 -3^x\ln x \ln 2}{2^x\cos x}\right)+\left(\dfrac{3^x+3^x\cdot x\tan x \ln x}{2^x x\cos x}\right)\)

=\(\dfrac{3^x(x\ln x \ln 3 - x \ln x \ln 2+1+x\tan x \ln x)}{2^x x \cos x}\)

I will name this fraction y.

Then we find the derivative of Part1,2:

\(\dfrac{3^x \ln x(x\ln x \ln 3 - x \ln x \ln 2 + 1 + x \tan x \ln x)+3^x(\ln x \ln 3 + \ln 3 -\ln x \ln 2 - \ln 2+\tan x \ln x+x\sec^2x\ln x+\tan x)}{2^{2x}x^2\cos^2x}\)Oops it's too long :( I will name this looooooonnnnnggggg fraction z then)

Then we use product rule for Part1,2 and Part 3:

f'(x)= \(y(\dfrac{x^{\sqrt3-2}+(\sqrt3-1)(\ln x)(x^{\sqrt3-2})}{2\sqrt{\ln x(x^{\sqrt3-1})}})+z(\sqrt{\ln x(x^{\sqrt3-1})})\)