Question

Find dy/dx if y=x/√1+x²

Answer 1

Answer:

$dy/dx= 1/(1+x^2)^{3/2}$

Step-by-step explanation:

$y=x/\sqrt{1+x^2}$

∵ Quotient rule states

$d(u/v)/dx = (v.du/dx-udv/dx)/v^2$

∴ $d(x/\sqrt{1+x^2})/dx=[\sqrt{1+x^2}.d(x)/dx-x.d(\sqrt{1+x^2})]/ (\sqrt{1+x^2} )^2$

$d(x)/dx=1$ and $d(\sqrt{1+x^2}) /dx=x/\sqrt{1+x^2}$ {By chain rule}

∴ $dy/dx=[\sqrt{1+x^2} -(x^2/\sqrt{1+x^2})]/1+x^2$

⇒ $dy/dx=(1+x^2-x^2) / (\sqrt{1+x^2})(1+x^2)$

⇒ $dy/dx= 1/(1+x^2)^{3/2}$