What is the derivative of y = (xsqrt(x^2 + 1))/(x + 1)^(2/3)?

1 Answer
Aug 23, 2017

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

Explanation:

We let

y = (xsqrt(x^2 + 1))/(x + 1)^(2/3)

Now using logarithmic differentiation, we have:

lny = ln((xsqrt(x^2 + 1))/(x + 1)^(2/3))

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) + 1/2ln(x^2 + 1) - 2/3ln(x + 1)

Now differentiate term by term .

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

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

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

Hopefully this helps!