Question

use logarithmic differentiation to find dy/dx y= ((x+2)sqrt(1-x^2))/4x^3 $y= ((x+2) \sqrt(1-x^2))/4x^3$

Answer 1

Answer: dy/dx = (x+2)(x^3-3/2(1-x^2)^1/2 x)+(1-x^2)^1/2/2x^3/2

Step-by-step explanation: Differentiate each term with respect to x, then solve for y.

I hope this helps you out!

Answer 2

If

$y = \dfrac{(x+2)\sqrt{1-x^2}}{4x^3}$

then taking the logarithm of both sides and expanding the right side gives

$\ln(y) = \ln\left(\dfrac{(x+2)\sqrt{1-x^2}}{4x^3}\right) \\\\ \ln(y) = \ln(x+2) + \ln\left(\sqrt{1-x^2}\right) - \ln(4) - \ln(x^3) \\\\ \ln(y) = \ln(x+2) + \dfrac12 \ln\left(1-x^2\right) - \ln(4) - 3\ln(x)$

Differentiate both sides with respect to x :

$\dfrac1y \dfrac{\mathrm dy}{\mathrm dx} = \dfrac1{x+2} + \dfrac{-2x}{2(1-x^2)} - 0 - \dfrac3x \\\\ \dfrac 1y \dfrac{\mathrm dy}{\mathrm dx} = \dfrac1{x+2} - \dfrac{x}{1-x^2} - \dfrac3x \\\\ \dfrac 1y \dfrac{\mathrm dy}{\mathrm dx} = -\dfrac{x^3+4x^2-2x-6}{x(x+2)(1-x^2)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = y \left(-\dfrac{x^3+4x^2-2x-6}{x(x+2)(1-x^2)}\right) \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{(x+2)\sqrt{1-x^2}}{4x^3}\left(-\dfrac{x^3+4x^2-2x-6}{x(x+2)(1-x^2)}\right) \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \boxed{-\dfrac{x^3+4x^2-2x-6}{4x^4\sqrt{1-x^2}}}$