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

1 Answer
Noah G
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!

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