Differential Calculus> differentiate tan^-1( sqrt(1+x^2) – sqrt(...

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

taniska , 10 Years ago

Grade 12

1 Answers

Jitender Singh

Last Activity: 10 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
$\frac{dy}{dx} = \frac{dtan^{-1}u}{dx}.\frac{du}{dx}$
$u = \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}$
$u^{2} = \frac{2-2\sqrt{1-x^{4}}}{2+2\sqrt{1+x^{4}}}$
$u^{2} = \frac{1-\sqrt{1-x^{4}}}{1+\sqrt{1+x^{4}}}$
$1 + u^{2} = 1 + \frac{1-\sqrt{1-x^{4}}}{1+\sqrt{1+x^{4}}}$
$1 + u^{2} = \frac{2}{1+\sqrt{1+x^{4}}}$
$\frac{dy}{dx} = \frac{dtan^{-1}u}{dx}.\frac{du}{dx}$
$\frac{dy}{dx} = \frac{1}{1+u^{2}}.\frac{du}{dx}$
$u = \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}$
$u = \frac{\sqrt{1+x^{2}}-\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}.\frac{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}{\sqrt{1+x^{2}}+\sqrt{1-x^{2}}}$
$u = \frac{x^{4}}{1+\sqrt{1-x^{4}}}$
$\frac{dy}{dx} = \frac{1+\sqrt{1+x^{4}}}{2}.\frac{(1+\sqrt{1-x^{4}}).4x^{3}-x^{4}(\frac{-4x^{3}}{2\sqrt{1-x^{4}}})}{(1+\sqrt{1-x^{4}})^{2}}$
$\frac{dy}{dx} = \frac{1+\sqrt{1+x^{4}}}{2}.\frac{(1+\sqrt{1-x^{4}}).4x^{3}+(\frac{2x^{7}}{\sqrt{1-x^{4}}})}{(1+\sqrt{1-x^{4}})^{2}}$
$\frac{dy}{dx} = \frac{1+\sqrt{1+x^{4}}}{2}.\frac{(1-x^{4}+\sqrt{1-x^{4}}).4x^{3}+2x^{7}}{(1+\sqrt{1-x^{4}})^{2}.\sqrt{1-x^{4}}}$