- If y=tan⁡−1(bx−αax+b)+tan⁡−1(θ2−θ11+θ2θ1)+⋯+tan⁡−1(θn−θn−11+θnθn−1)−tan⁡−1(θn)y = \tan^{-1} \left( \frac{bx - \alpha}{ax + b} \right) + \tan^{-1} \left( \frac{\theta_2 - \theta_1}{1 + \theta_2 \theta_1} \right) + \dots + \tan^{-1} \left( \frac{\theta_n - \theta_{n-1}}{1 + \theta_n \theta_{n-1}} \right) - \tan^{-1}(\theta_n)…

Mathematics

Problem

- If y=tan⁡−1(bx−αax+b)+tan⁡−1(θ2−θ11+θ2θ1)+⋯+tan⁡−1(θn−θn−11+θnθn−1)−tan⁡−1(θn)y = \tan^{-1} \left( \frac{bx - \alpha}{ax + b} \right) + \tan^{-1} \left( \frac{\theta_2 - \theta_1}{1 + \theta_2 \theta_1} \right) + \dots + \tan^{-1} \left( \frac{\theta_n - \theta_{n-1}}{1 + \theta_n \theta_{n-1}} \right) - \tan^{-1}(\theta_n) , then find dydx\frac{dy}{dx} .

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