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differentiate tan^-1( sqrt(1+x^2) – sqrt(1-x^2) / sqrt(1+x^2) + s qrt(1-x^2))

differentiate tan^-1(sqrt(1+x^2) – sqrt(1-x^2) / sqrt(1+x^2) + sqrt(1-x^2))

Grade:12

1 Answers

Jitender Singh IIT Delhi
askIITians Faculty 158 Points
9 years ago
Ans:
Hello Student,
Please find answer to your question below

y = tan^{-1}(\frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}})
Apply the chain rule
\frac{dy}{dx} = \frac{dtan^{-1}u}{dx}.\frac{du}{dx}
u = \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}
u^{2} = \frac{2-2\sqrt{1-x^{4}}}{2+2\sqrt{1+x^{4}}}
u^{2} = \frac{1-\sqrt{1-x^{4}}}{1+\sqrt{1+x^{4}}}
1 + u^{2} = 1 + \frac{1-\sqrt{1-x^{4}}}{1+\sqrt{1+x^{4}}}
1 + u^{2} = \frac{2}{1+\sqrt{1+x^{4}}}
\frac{dy}{dx} = \frac{dtan^{-1}u}{dx}.\frac{du}{dx}
\frac{dy}{dx} = \frac{1}{1+u^{2}}.\frac{du}{dx}
u = \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}
u = \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}.\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}
u = \frac{x^{4}}{1+\sqrt{1-x^{4}}}
\frac{dy}{dx} = \frac{1+\sqrt{1+x^{4}}}{2}.\frac{(1+\sqrt{1-x^{4}}).4x^{3}-x^{4}(\frac{-4x^{3}}{2\sqrt{1-x^{4}}})}{(1+\sqrt{1-x^{4}})^{2}}
\frac{dy}{dx} = \frac{1+\sqrt{1+x^{4}}}{2}.\frac{(1+\sqrt{1-x^{4}}).4x^{3}+(\frac{2x^{7}}{\sqrt{1-x^{4}}})}{(1+\sqrt{1-x^{4}})^{2}}
\frac{dy}{dx} = \frac{1+\sqrt{1+x^{4}}}{2}.\frac{(1-x^{4}+\sqrt{1-x^{4}}).4x^{3}+2x^{7}}{(1+\sqrt{1-x^{4}})^{2}.\sqrt{1-x^{4}}}



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