What is the derivative of #g(u) = ln(sqrt((3u+6)/(3u-6)))#?

Answer 1

#g^' = -2 * 1/((u+2)(u-2))#

First, notice that you can simplify the function o get

#g(u) = ln(sqrt((color(red)(cancel(color(black)(3)))(u+2))/(color(red)(cancel(color(black)(3)))(u-2)))) = ln(sqrt((u+2)/(u-2)))#

You can differentiate this function by using the chain rule twice and the quotient rule once.

More specifically, you'll need to use the chain rule for

So, start with the target derivative

#d/(du)(g) = d/(dt)lnt * d/(du)(t)#
#g^' = 1/t * d/(du)(sqrt((u+2)/(u-2)))#
#g^' = 1/t * d/(dv)sqrt(v) * d/(du)(v)#
#g^' = 1/t * 1/2 * 1/sqrt(v) * d/(du)((u+2)/(u-2))#

This is where the quotient rule comes in handy

#d/(du)((u+2)/(u-2)) = ([d/dx(u+2)] * (u-2) - (u+2) * d/dx(u-2))/(u-2)^2#
#d/(du)((u+2)/(u-2)) = (1 * (u-2) - (u+2) * 1)/(u-2)^2#
#d/(du)((u+2)/(u-2)) = (color(red)(cancel(color(black)(u))) - 2 - color(red)(cancel(color(black)(u))) - 2)/(u-2)^2#
#d/(du)((u+2)/(u-2)) = -4/(u-2)^2#

This means that the target derivative will be equal to

#g^' = 1/2 * 1/sqrt((u+2)/(u-2)) * 1/sqrt((u+2)/(u-2)) * (-4/(u-2)^2)#
#g^' = - 2 * (1/sqrt((u+2)/(u-2)))^2 * 1/(u-2)^2#
#g^' = -2 * color(red)(cancel(color(black)(u-2)))/(u+2) * 1/(u-2)^color(red)(cancel(color(black)(2)))#
#g^' = color(green)(-2 * 1/((u+2)(u-2)))#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 2

#g'(u) = (-2)/((u+2)(u-2)) = (-2)/(u^2-4)#

Factor and reduce the innermost fraction, then use properties of logarithms to rewrite #g#
#g(u) = ln(sqrt((3u+6)/(3u-6)))= ln(sqrt((3(u+2))/(3(u-2))))#
# = ln(sqrt((u+2)/(u-2)))#
# = 1/2ln((u+2)/(u-2))#
# = 1/2[ln(u+2)-ln(u-2)]#
Now differentiate with respect to #u#:
#g'(u) = 1/2[1/(u+2)-1/(u-2)]#
# = 1/2[((u-2)-(u+2))/((u+2)(u-2))]#
# = 1/2[(-4)/((u+2)(u-2))]#
# = (-2)/((u+2)(u-2)) = (-2)/(u^2-4)#
Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer 3

To find the derivative of ( g(u) = \ln\left(\sqrt{\frac{3u+6}{3u-6}}\right) ), we use the chain rule and the properties of logarithmic and square root functions. The derivative is:

[ g'(u) = \frac{d}{du} \ln\left(\sqrt{\frac{3u+6}{3u-6}}\right) ] [ = \frac{1}{\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{d}{du} \left(\frac{3u+6}{3u-6}\right)^{\frac{1}{2}} ] [ = \frac{1}{\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{1}{2} \cdot \frac{d}{du} \left(\frac{3u+6}{3u-6}\right)^\frac{1}{2} ] [ = \frac{1}{\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{1}{2} \cdot \frac{1}{\left(\frac{3u+6}{3u-6}\right)^\frac{1}{2}} \cdot \frac{d}{du} \left(\frac{3u+6}{3u-6}\right) ] [ = \frac{1}{\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{1}{2} \cdot \frac{1}{\left(\frac{3u+6}{3u-6}\right)^\frac{1}{2}} \cdot \frac{d}{du} \left(\frac{3u+6}{3u-6}\right) ] [ = \frac{1}{\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{d}{du} \left(\frac{3u+6}{3u-6}\right) ] [ = \frac{1}{2\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{d}{du} \left(\frac{3u+6}{3u-6}\right) ] [ = \frac{1}{2\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{d}{du} \left(\frac{(3u+6)^{\frac{1}{2}}}{(3u-6)^{\frac{1}{2}}}\right) ] [ = \frac{1}{2\sqrt{\frac{3u+6}{3u-6}}} \cdot \left(\frac{\frac{1}{2}(3u+6)^{-\frac{1}{2}}}{(3u-6)^{\frac{1}{2}}} - \frac{\frac{1}{2}(3u-6)^{-\frac{1}{2}}}{(3u+6)^{\frac{1}{2}}}\right) ] [ = \frac{1}{2\sqrt{\frac{3u+6}{3u-6}}} \cdot \left(\frac{3u+6}{2(3u-6)} - \frac{3u-6}{2(3u+6)}\right) ] [ = \frac{1}{2\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{3u+6 - (3u-6)}{2(3u-6)(3u+6)} ] [ = \frac{1}{2\sqrt{\frac{3u+6}{3u-6}}} \cdot \frac{12}{(3u-6)(3u+6)} ] [ = \frac{6}{(3u-6)(3u+6)\sqrt{\frac{3u+6}{3u-6}}} ]

So, the derivative of ( g(u) ) is ( \frac{6}{(3u-6)(3u+6)\sqrt{\frac{3u+6}{3u-6}}} ).

Sign up to view the whole answer

By signing up, you agree to our Terms of Service and Privacy Policy

Sign up with email
Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

Not the question you need?

Drag image here or click to upload

Or press Ctrl + V to paste
Answer Background
HIX Tutor
Solve ANY homework problem with a smart AI
  • 98% accuracy study help
  • Covers math, physics, chemistry, biology, and more
  • Step-by-step, in-depth guides
  • Readily available 24/7