Question

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

Answer 1

`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!