Show that: $\frac{\mathrm{d}}{\mathrm{d}x}\sin^{-1}x = \frac{1}{\sqrt{1 - x^{2}}}$ \(\frac{\mathrm

Question

Show that: $\frac{\mathrm{d}}{\mathrm{d}x}\sin^{-1}x = \frac{1}{\sqrt{1 - x^{2}}}$ $\frac{\mathrm{d}}{\mathrm{d}x}{{\cos }^{-1}}x=-\frac{1}{\sqrt{1-{{x}^{2}}}}$ $\frac{\mathrm{d}}{\mathrm{d}x}\tan^{-1}x = \frac{1}{1 + x^{2}}$

Accepted Answer

(i) $\frac1{\sqrt{1-x^2}}$ (ii) $-\frac1{\sqrt{1-x^2}}$ (iii) $\frac1{1+x^2}$

Techniques of Differentiation - Deriving Inverse Trigonometric Derivatives

Techniques of Differentiation - Deriving Inverse Trigonometric Derivatives

Video Solution:

Solution:

Solution locked