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Mathematics Question on Differential equations

Find the differentiation of cot−13+4tan x4−3tan xcot^{-1}\frac{3+4tan\,x}{4-3tan\,x}

Answer

To find the derivative of cot−13+4tan x4−3tan xcot^{-1}\frac{3+4tan\,x}{4-3tan\,x}, we can use the chain rule and the derivative of the inverse cotangent function.
Let u = (3+4tan x)(4−3tan x)\frac{(3+4tan\,x)}{(4-3tan\,x)}
Then, cot−13+4tan x4−3tan x=cot−1(u)cot^{-1}\frac{3+4tan\,x}{4-3tan\,x}=cot^{-1}(u)
Differentiating both sides with respect to x, we get:
ddx[cot−1(u)]=ddx[cot−1(3+4tan x)(4−3tan x)]\frac{d}{dx}[cot^{-1}(u)]=\frac{d}{dx}[cot^{-1}\frac{(3+4tan\,x)}{(4-3tan\,x)}]
Using the chain rule, we have:
ddx[cot−1(u)]=−1[1+u2]×dudx\frac{d}{dx}[cot^{-1}(u)]=\frac{-1}{[1+u^2]}\times\frac{du}{dx}
To find du/dx, we can use the quotient rule:
dudx=[(4−3tan x)(4sec2 x)−(3+4 tan x)(3 sec2x)](4−3 tan x)2\frac{du}{dx}=\frac{[(4-3tan\,x)(4sec^2\,x)-(3+4\,tan\,x)(3\,sec^2x)]}{(4-3\,tan\,x)^2}
Simplifying this expression, we get:
dudx=−48 sec2(4−3 tan x)2\frac{du}{dx}=\frac{-48\,sec^2}{(4-3\,tan\,x)^2}
Substituting this expression for dudx\frac{du}{dx} into our earlier equation, we get:
ddx[cot−1(3+4tan x)(4−3tan x)]=−1[1+u2]×[−48 sec2 x(4−3tan x)2]\frac{d}{dx}[cot^{-1}\frac{(3+4tan\,x)}{(4-3tan\,x)}]=\frac{-1}{[1+u^2]}\times[\frac{-48\,sec^2\,x}{(4-3tan\,x)^2}]
Simplifying this expression, we get:
ddx[cot−1(3+4tan x)(4−3tan x)]=48sec2 x[(3tan x−4)2+25]\frac{d}{dx}[cot^{-1}\frac{(3+4tan\,x)}{(4-3tan\,x)}]=\frac{48sec^2\,x}{[(3tan\,x-4)^2+25]}
Therefore, the derivative of [cot−1(3+4tan x)(4−3tan x)][cot^{-1}\frac{(3+4tan\,x)}{(4-3tan\,x)}] is 48sec2 x[(3tan x−4)2+25]\frac{48sec^2\,x}{[(3tan\,x-4)^2+25]}.