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differentiate the following function with respect to x
tan^-1(x / a+sqrt(a^2 – x^2)) , -a

taniska, 7 years ago

Grade:12

1 Answers

Arun Kumar IIT Delhi

askIITians Faculty 256 Points

7 years ago

Hi

$\\\mathrm{Let\:}\frac{x}{a+\sqrt{a^2-x^2}}\mathrm{=}u \\=\frac{d}{du}\left(\arctan \left(u\right)\right)\frac{d}{dx}\left(\frac{x}{a+\sqrt{a^2-x^2}}\right) \\\frac{d}{dx}\left(\frac{x}{a+\sqrt{a^2-x^2}}\right) \\=\frac{\frac{d}{dx}\left(x\right)\left(a+\sqrt{a^2-x^2}\right)-\frac{d}{dx}\left(a+\sqrt{a^2-x^2}\right)x}{\left(a+\sqrt{a^2-x^2}\right)^2} \\=\frac{1\left(a+\sqrt{a^2-x^2}\right)-\left(-\frac{x}{\sqrt{a^2-x^2}}\right)x}{\left(a+\sqrt{a^2-x^2}\right)^2} \\=\frac{1}{-\frac{x^2}{a}+\sqrt{a^2-x^2}+a} \\=>\frac{1}{u^2+1}\frac{1}{-\frac{x^2}{a}+\sqrt{a^2-x^2}+a} \\\mathrm{Substitute\:}\:u=\frac{x}{a+\sqrt{a^2-x^2}} \\=\frac{1}{\left(\frac{x}{a+\sqrt{a^2-x^2}}\right)^2+1}\frac{1}{-\frac{x^2}{a}+\sqrt{a^2-x^2}+a} \\=\frac{1}{2\sqrt{a^2-x^2}}$

Thanks & Regards, Arun Kumar, Btech,IIT Delhi, Askiitians Faculty

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