Question

Find the differentiation of \(cot^{-1}\frac{3+4tan\,x}{4-3tan\,x}\)

Answer 1

To find the derivative of \( \cot^{-1} \left( \frac{3+4\tan{x}}{4-3\tan{x}} \right) \), we use the chain rule and the derivative of the inverse cotangent function.
Let \[ u = \frac{3+4\tan{x}}{4-3\tan{x}} \]
Then, \[ \cot^{-1} \left( \frac{3+4\tan{x}}{4-3\tan{x}} \right) = \cot^{-1}(u) \]
Differentiating both sides with respect to \( x \), we get: \[ \frac{d}{dx} \left[ \cot^{-1}(u) \right] = \frac{d}{dx} \left[ \cot^{-1} \left( \frac{3+4\tan{x}}{4-3\tan{x}} \right) \right] \]
Using the chain rule: \[ \frac{d}{dx} \left[ \cot^{-1}(u) \right] = \frac{-1}{1+u^2} \times \frac{du}{dx} \] To find \( \frac{du}{dx} \), we apply the quotient rule: \[ \frac{du}{dx} = \frac{(4-3\tan{x})(4\sec^2{x}) - (3+4\tan{x})(3\sec^2{x})}{(4-3\tan{x})^2} \] Simplifying this: \[ \frac{du}{dx} = \frac{-48\sec^2{x}}{(4-3\tan{x})^2} \] Substituting into the earlier equation: \[ \frac{d}{dx} \left[ \cot^{-1} \left( \frac{3+4\tan{x}}{4-3\tan{x}} \right) \right] = \frac{-1}{1+u^2} \times \frac{-48\sec^2{x}}{(4-3\tan{x})^2} \] Simplifying the expression: \[ \frac{d}{dx} \left[ \cot^{-1} \left( \frac{3+4\tan{x}}{4-3\tan{x}} \right) \right] = \frac{48\sec^2{x}}{( (3\tan{x} - 4)^2 + 25 )} \] Therefore, the derivative is: \[ \frac{48\sec^2{x}}{( (3\tan{x} - 4)^2 + 25 )} \]