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

If

`y = ((x+2)√(1-x^2))/(4x^3)`

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

`\ln(y) = \ln\left(((x+2)√(1-x^2))/(4x^3)\right) \\\\ \ln(y) = \ln(x+2) + \ln\left(√(1-x^2)\right) - \ln(4) - \ln(x^3) \\\\ \ln(y) = \ln(x+2) + \frac12 \ln\left(1-x^2\right) - \ln(4) - 3\ln(x)`

Differentiate both sides with respect to x :

`\frac1y (\mathrm dy)/(\mathrm dx) = \frac1{x+2} + (-2x)/(2(1-x^2)) - 0 - \frac3x \\\\ \frac 1y (\mathrm dy)/(\mathrm dx) = \frac1{x+2} - (x)/(1-x^2) - \frac3x \\\\ \frac 1y (\mathrm dy)/(\mathrm dx) = -(x^3+4x^2-2x-6)/(x(x+2)(1-x^2)) \\\\ (\mathrm dy)/(\mathrm dx) = y \left(-(x^3+4x^2-2x-6)/(x(x+2)(1-x^2))\right) \\\\ (\mathrm dy)/(\mathrm dx) = ((x+2)√(1-x^2))/(4x^3)\left(-(x^3+4x^2-2x-6)/(x(x+2)(1-x^2))\right) \\\\ (\mathrm dy)/(\mathrm dx) = \boxed{-(x^3+4x^2-2x-6)/(4x^4√(1-x^2))}`

Answer 2

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!