Question

The differential coefficient of $co{\sec }^{-1}\frac{1}{2x^2-1}$ w.r.t $\sqrt{1-x^2}$ at $x=\frac{1}{2}$ is equal to

Answer 1

$dcose{c}^{-1}\frac{1}{2x^2-1}$

$\overline{d}\sqrt{1-x^2}$

$\frac{d}{dx}cose{c}^{-1}(1/2x^2+1).\frac{dx}{d\sqrt{1-x^2}}$

${\frac{1}{\left(\frac{1}{2x^2+1}\right)\sqrt{1-\frac{1}{(2x^2-1{)}^2}}}}^{\left(4x\right)}.\frac{2\sqrt{1-x^2}}{1(1-2x)}$

$x=1/2\frac{1}{\left(2\right(\frac{1}{x})-1)\sqrt{1-\frac{1}{\left(2\right(\frac{1}{x})-1{)}^2}}}.\frac{4}{\left(2\right(1/2{)}^2-1{)}^2}.\sqrt{1(1/2{)}^2}$

$\frac{1}{(1/2-1)\sqrt{1-\frac{1}{x}}}\frac{4}{\left(1/4\right)}\frac{\sqrt{3}}{2}$

$\frac{-1}{\left(1/2\right)}\frac{2}{\sqrt{3}}\frac{4}{\left(1/4\right)}\frac{\sqrt{3}}{2}$

$-32$