What is the derivative of y = (xsqrt(x^2 + 1))/(x + 1)^(2/3)y=x√x2+1(x+1)23?
1 Answer
Explanation:
We let
y = (xsqrt(x^2 + 1))/(x + 1)^(2/3)y=x√x2+1(x+1)23
Now using logarithmic differentiation, we have:
lny = ln((xsqrt(x^2 + 1))/(x + 1)^(2/3))lny=ln(x√x2+1(x+1)23)
We can now use laws of logarithms to simplify.
lny = ln(x) + ln(x^2 + 1)^(1/2) - ln(x + 1)^(2/3)lny=ln(x)+ln(x2+1)12−ln(x+1)23
lny = ln(x) + 1/2ln(x^2 + 1) - 2/3ln(x + 1)lny=ln(x)+12ln(x2+1)−23ln(x+1)
Now differentiate term by term .
1/y(dy/dx) = 1/x + (2x)/(2(x^2 + 1)) - 2/(3(x+ 1))1y(dydx)=1x+2x2(x2+1)−23(x+1)
dy/dx = y(1/x + x/(x^2 + 1) - 2/(3(x + 1)))dydx=y(1x+xx2+1−23(x+1))
dy/dx = (xsqrt(x^2 + 1))/(x + 1)^(2/3)(1/x + x/(x^2 + 1) - 2/(3(x + 1)))dydx=x√x2+1(x+1)23(1x+xx2+1−23(x+1))
Hopefully this helps!
