# Differentiate cos^-1(2x^2-1) with respect to sin^-1√1-x^2?

Jan 22, 2018

$\frac{\mathrm{df}}{\mathrm{dg}} = \sqrt{\frac{2 - {x}^{2}}{1 - {x}^{2}}}$

#### Explanation:

Let $f \left(x\right) = {\cos}^{- 1} \left(2 {x}^{2} - 1\right)$ and $g \left(x\right) = {\sin}^{- 1} \left(1 - {x}^{2}\right)$

what we are seeking is $\frac{\mathrm{df}}{\mathrm{dg}}$, which is equal to $\frac{\frac{\mathrm{df}}{\mathrm{dx}}}{\frac{\mathrm{dg}}{\mathrm{dx}}}$

As $f = {\cos}^{- 1} \left(2 {x}^{2} - 1\right)$, we have $\cos f = 2 {x}^{2} - 1$

and differentiating $- \sin f \cdot \frac{\mathrm{df}}{\mathrm{dx}} = 4 x$

or $\frac{\mathrm{df}}{\mathrm{dx}} = - \frac{4 x}{\sin} f = - \frac{4 x}{\sqrt{1 - {\cos}^{2} f}} = - \frac{4 x}{\sqrt{1 - \left(2 {x}^{2} - 1\right)}} ^ 2$

= $- \frac{4 x}{\sqrt{1 - 4 {x}^{4} + 4 {x}^{2} - 1}} = - \frac{4 x}{2 x \sqrt{1 - {x}^{2}}} = - \frac{2}{\sqrt{1 - {x}^{2}}}$

Similarly we have $\sin g = 1 - {x}^{2}$ and $\cos g \cdot \frac{\mathrm{dg}}{\mathrm{dx}} = - 2 x$

or $\frac{\mathrm{dg}}{\mathrm{dx}} = - \frac{2 x}{\cos} g = - \frac{2 x}{\sqrt{1 - {\sin}^{2} g}} = - \frac{2 x}{\sqrt{1 - {\left(1 - {x}^{2}\right)}^{2}}}$

= $- \frac{2 x}{\sqrt{1 - 1 + 2 {x}^{2} - {x}^{4}}} = - \frac{2}{\sqrt{2 - {x}^{2}}}$

Hence $\frac{\mathrm{df}}{\mathrm{dg}} = \frac{- \frac{2}{\sqrt{1 - {x}^{2}}}}{- \frac{2}{\sqrt{2 - {x}^{2}}}}$

= $\sqrt{\frac{2 - {x}^{2}}{1 - {x}^{2}}}$