derivative of natural log function?

Let's make life easy and apply the log laws. Check out this piece of magic.

G(u) = ln [(3u+6)/(3u-6)]^(1/2)

= 1/2 * ln [(3u+6)/(3u-6)]

= 1/2 * [ln(3u+6) - ln(3u-6) ]

Now differentiating

G'(u) = 1/2 * [ 3/(3u+6) - 3 / (3u-6) ]

= 1/2 * [1/(u+2) - 1/(u-2) ]

You could leave it like that, or get a common denom

G' = -2 / (u^2 - 4)

= 2 / (4 - u^2)

Before differentiating this expression, a bit of simplification is in order. (I am also assuming that the missing closing parenthesis is at the end of the expression.)

G(u) = ln sqrt ((3u+6)/(3u-6))

= ln (((3u+6) / (3u-6)) ^ (1/2))

Using the power of logs, this becomes

(1/2) * ln ((3u+6)/(3u-6))

Cancelling the common factor of 3 in the numerator and denominator of the ln argument:

= (1/2) * ln ( (u+2) / (u-2) )

We may now break apart the quotient as two separate ln functions:

= (1/2) * (ln (u+2) - ln (u-2))

Now we can differentiate easily, giving

G'(u) = (1/2) * ( 1 / (u + 2) - 1 / (u - 2) )

which can be further simplified to

(1/2) * (((u-2) - (u+2)) / (u^2 - 4))

= (-4/2) / (u^2 - 4)

= -2 / (u^2 - 4)

Just a couple applications of the chain rule...

G(u) = ln sqrt((3u + 6)/(3u - 6)

= ln [(3u + 6)/(3u - 6)]^(1/2)

The chain rule says to first take the derivative of the outter function with the inner function left alone. The outter function is the natural log. So if we apply the chain rule, we get...

1/[(3u + 6)/(3u - 6)]^(1/2)

Now we need to multiply this by the derivative of the inner function. The thing is, the inner function itself is a composition of functions so we need to apply the chain rule again.

[(3u + 6)/(3u - 6)]^(1/2)

The outter function is the square root function. So we need to take the derivative of this, with the inner function left alone. Then multiply it by the derivative of the inner function (but for that, we need to use the quotient rule). So let's do that...

1/[(2)sqrt((3u + 6)/(3u - 6))] * [(3)(3u - 6) - (3)(3u + 6)]/(3u -6)^2

So this whole mess gets multiplied by the derivative we took earlier of the natural log function.

G'(u) = [1/[(3u + 6)/(3u - 6)]^(1/2)] * 1/[(2)sqrt((3u + 6)/(3u - 6))] * [(3)(3u - 6) - (3)(3u + 6)]/(3u -6)^2

G'(u) = ln ([(3)/(3u - 6)] - [((3)(3u + 6))/(3u - 6)^2]) / (2)((3u + 6)/(3u - 6))^(1/2)

It's not very nice looking. There are a few algebraic simplifications you can make but it's still going to look pretty bad.

The best way to make it look neatest would be to apply the log rules first like the answerers above did. But either way can be used and you'll get the same result.

G(u) = (1/2) [ ln(3u + 6) -ln (3u - 6)]

d/du ln x = 1/x

dG(u)/du = (1/2) [ (1/(3u+6))*3 - 3/(3u-6)]

dG(u)/du = (1/2) [ (1/(u+2) - 1/(u-2)]

= (1/2) [(u-2-u-2) / (u^2 - 4)

= -2/(u^2 -4)

At least you have a selection of answers to choose from!

G(u) = ln sqrt((3u+6)/(3u-6)

Remember the rules of logs:

log(a/b) = log(a)-log(b)

log(sqrt(a)) = (1/2)log(a)

G(u) = (1/2)[ln(3u+6) - ln(3u-6)]

G'(u) = (1/2)[3/(3u+6) - 3/(3u-6)]

=(1/2)[1/(u+2) - 1/(u-2)]

When you differentiate the natural log of a function ln|f(x)| You get 1 over the function, multiplied by the derivative of that function: f'(x)/f(x) Hope that helps!

[1(sqrt((3u+6)/(3u-6))] * [1/(2*sqrt((3u+6)/(3u-6)))] * [((3u-6)*3 - (3u+6)*3) / ((3u-6)^2)]

chain rule or do the log rule first like the smart people above me did

you do the simplifying

G(u) = ln((3u+6)/(3u-6))^½

= ½ ln((3u+6)/(3u-6))

= ½ [ln(3u+6) - ln(3u-6)]

G '(u) = ½ [3/(3u + 6) - 3/(3u - 6)]

......... = ½ [1/(u + 2) - 1/(u - 2)]

......... = ½ [ -4/(u + 2)(u - 2)]

......... = -2/[(u + 2)(u - 2)]

you left out an unbalanced bracket in your question so here are the two variations and their answers:

ln ( sqrt((3*u+6)/(3*u-6) )

1/2/(3*u+6)*(3*u-6)*(3/(3*u-6)-3*(3*u+6)/(3*u-6)^2)

ln ( sqrt(3*u+6)) /(3*u-6) )

3/2/(3*u+6)/(3*u-6)-3*ln((3*u+6)^(1/2))/(3*u-6)^2

18u/2(9u^2-36)

I think..

here is my logic..

you have to use the chain rule,,

the derivative of the outside function, times the derivative of the inside function, times derivative of u..

yea, that does it..