Question

\( x\left(x^{2}+1\right)^{-1 / 2} \) find dy/dx using chain rule

Answer 1

\(\frac d{dx} x(x^2 + 1)^{\frac{-1}2} = x\frac d{dx} (x^2 + 1)^{\frac{-1}2} + \frac{dx}{dx} \times (x^2 + 1)^{\frac{-1}2}\)

\( = x\times \frac{-1}2 (x^2 + 1)^{\frac{-1}2-1} \frac{d}{dx} (x^2 + 1) + (x^2 + 1)^{\frac{-1}2}\)

\(= -\frac x2 (x^2 + 1)^{\frac{-3}2}.2x + (x^2 + 1)^\frac{-1}2\)

\( = (x^2 + 1)^\frac{-1}2 \left(\frac{-x^2}{x^2 + 1}+ 1\right)\)

\( = (x^2 + 1)^\frac{-1}2 \left(\frac1{x^2 + 1}\right)\)

\( = \frac1{(x^2 + 1)^\frac32}\)