What is the derivative of #ln[sqrt((2+x^2)/(2-x^2))]#?

Question

Emma Aldridge · Accepted Answer

#y^' = (4x)/((2-x^2) * (2 + x^2))#

You're going to have to use a combination of derivative rules to differentiate this function.

Right from the start, it's clear that you'll use the chain rule twice, for

At that point, you can use the quotient rule to complete the integration.

So, the derivative of #y# will be

#d/dx(y) = d/(du) * ln(u) * d/dx(underbrace(sqrt((2+x^2)/(2-x^2)))_(color(blue)(z)))#

Now focus on finding #d/dx(color(blue)(z))#, which will be equal to

#d/dx(sqrt(u_1)) = [d/(du_1)sqrt(u_1)] * d/dx(underbrace((2+x^2)/(2-x^2))_(color(orange)(w)))#

This time, use the quotient rule to find #d/dx(color(orange)(w))#

#d/dx(w) = ([d/dx(2+x^2)] * (2 - x^2) - (2 + x^2) * d/dx(2 - x^2))/(2-x^2)^2#

#w^' = [2x(2-x^2) - (2 + x^2) * (-2x)]/(2-x^2)^2#

#w^' = (4x - color(red)(cancel(color(black)(2x^3))) + 4x + color(red)(cancel(color(black)(2x^3))))/(2-x^2)^2#

#w^' = (8x)/(2-x^2)^2#

Plug this into the calculation of #d/dx(color(blue)(z))# to get

#d/dx(sqrt(u_1)) = 1/2u_1^(-1/2) * (8x)/(2 - x^2)^2#

#d/dx(sqrt((2+x^2)/(2-x^2))) = 1/2 * 1/(sqrt((2+x^2)/(2-x^2))) * (8x)/(2-x^2)^2#

Finally, plug this into the calculation for the target derivative to get

#d/dx(lnu) = 1/u * 1/2 * 1/(sqrt((2+x^2)/(2-x^2))) * (8x)/(2-x^2)^2#

#y^' = 1/sqrt((2+x^2)/(2-x)^2) * 1/2 * 1/(sqrt((2+x^2)/(2-x^2))) * (8x)/(2-x^2)^2#

#y^' = 1/2 * (1/sqrt((2+x^2)/(2-x^2)))^2 * (8x)/(2-x^2)^2#

#y^' = 1/2 * (2-x^2)/(2 + x^2) * (8x)/(2-x^2)^2#

This can be simplified to give

#y^' = 1/cancel(2) * color(red)(cancel(color(black)(2-x^2)))/(2 + x^2) * (cancel(8)^color(blue)(4)x)/(2-x^2)^color(red)(cancel(color(black)(2)))#

#y^' = color(green)((4x)/((2-x^2) * (2 + x^2)))#

Paisley Johnson · Answer

undefined The derivative of $ \ln\left(\sqrt{\frac{2+x^2}{2-x^2}}\right) $ is $ \frac{x}{x^4 - 4} $.