Solve:
${tan}^{- 1} (\frac{4 x}{1 + 5 x^{2}})$

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Solution

$\begin{matrix} {tan}^{- 1} (\frac{4 x}{1 + 5 x^{2}}) l e t x = tan θ, θ = {tan}^{- 1} x {tan}^{- 1} (\frac{5 x - x}{1 + 5 x \cdot x}) = {tan}^{- 1} (\frac{5 tan θ - tan θ}{1 + 5 {tan}^{2} θ}) {tan}^{- 1} tan (4 θ) [tan (A - B) = (\frac{tan A - tan B}{1 + tan A \cdot tan B})] [{tan}^{- 1} tan θ = θ] = 4 θ = 4 {tan}^{- 1} x A n s . \end{matrix}$

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{tan}^{- 1} (\frac{4 x}{1 + 5 x^{2}}) + {tan}^{- 1} (\frac{2 + 3 x}{3 - 2 x})

{tan}^{- 1} (\frac{4 x}{1 + 5 x^{2}}) + {tan}^{- 1} (\frac{2 + 3 x}{3 - 2 x})

5 x^{2} - 14 x + 9 = 0

\frac{d y}{d x}

y = {tan}^{- 1} (\frac{4 x}{1 + 5 x^{2}})

y = {tan}^{- 1} (\frac{4 x}{1 + 5 x^{2}}) + {tan}^{- 1} (\frac{2 + 3 x}{3 - 2 x})

\frac{d y}{d x}

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