Derivatives of logarithmic -

[939]

Derivatives of logarithmic functions - please help

Homework Statement

I am having trouble differentiating logarithmic functions. I am following this basic rule they gave us, namely: if y = ln g(x) then y' = g'(x)/g(x), but it is not working :(. Where am I going wrong?

Homework Equations

1) ln((4 - x)/(3x + 8))

2)ln(8x^3 - 3x)^1/2

The Attempt at a Solution

1)

(1)(3x + 8) - (3)(4 - x)
3x + 8 - 12 + 3x
...
(6x - 4/(3x + 8)^2)/(4 - x)(3x +8)

2)

(1/2(8x^3 - 3x)^-1/2)/((8x^2 - 3x)^1/2)

 

[InertialRef]

Homework Statement

I am having trouble differentiating logarithmic functions. I am following this basic rule they gave us, namely: if y = ln g(x) then y' = g'(x)/g(x), but it is not working :(. Where am I going wrong?

Homework Equations

1) ln((4 - x)/(3x + 8))

2)ln(8x^3 - 3x)^1/2

The Attempt at a Solution

1)

(1)(3x + 8) - (3)(4 - x)
3x + 8 - 12 + 3x
...
(6x - 4/(3x + 8)^2)/(4 - x)(3x +8)

2)

(1/2(8x^3 - 3x)^-1/2)/((8x^2 - 3x)^1/2)

You have the general equation of the derivative of the ln function wrong. The derivative of lng(x) = 1/g(x), not what you stated in your equation. Once you know that, it's a simple matter of using the chain rule to obtain the derivative.

 

[uart]

namely: if y = ln g(x) then y' = g'(x)/g(x), but it is not working :(.

That expression for y' is correct, however I think you are getting tripped up with either the calculation of g' or the subsequent algebraic simplification.

For (i), make sure you use the quotient rule for g' and make any appropriate cancellations of factors before simplifying further.

 

[Greger]

Maybe this might help,

Remember you can write the log as,
ln((4 - x)/(3x + 8)) = ln(4 - x) - ln(3x + 8)

Now when you take the derivative you can take it of each log separately so it will be a little easier to see.

For example, when taking the derivative of one, if it were

ln((100-x)/(1-x)) = ln(100 - x) - ln(1 - x) then:

d/dx(ln(100-x) = -1/(100-x) and d/dx(ln(1-x) = -1/(1-x)

(Comparing this to your formula, here in the first case g(x) = 100-x and g'(x) = -1)

So the answer derivative of ln((100-x)/(1-x)) with respect to x would be:

-1/(100-x) - (-1/(1-x)) = -1/(100-x) + 1/(1-x)

Hope that helps.

Sorry if you have already done this, I couldn't really follow your working.

 

[HallsofIvy]

Homework Statement

I am having trouble differentiating logarithmic functions. I am following this basic rule they gave us, namely: if y = ln g(x) then y' = g'(x)/g(x), but it is not working :(. Where am I going wrong?

Homework Equations

1) ln((4 - x)/(3x + 8))

2)ln(8x^3 - 3x)^1/2

This is ambiguous. Do you mean ln((x^3- 3x)^1/2) (which is equal to (1/2)ln(x^3- 3x)) or (ln(8x^3- 3x))^1/2?

The Attempt at a Solution

1)

(1)(3x + 8) - (3)(4 - x)[/quote]
The derivative of 4- x is -1, not 1

3x + 8 - 12 + 3x
...
(6x - 4/(3x + 8)^2)/(4 - x)(3x +8)

2)

(1/2(8x^3 - 3x)^-1/2)/((8x^2 - 3x)^1/2)

The derivative of (8x^3- 3x)^1/2 is (1/2)(x^3- 3x)^-1/2 times the derivative of 8x^3- 3x which is 3x^2- 3.