Derivative of ln[((x2+1).5) / x(2x3-1)2]

[lude1]

Homework Statement

Find f' of ln[((x²+1)^.5) / x(2x³-1)²]

Homework Equations

The Attempt at a Solution

I know the derivative of ln(x) is 1/x... however I don't know how to start this particular problem. I think there is a easier way to solve this without trying to solve the derivative of [((x²+1)^.5) / x(2x³-1)²]..?

 

[Saladsamurai]

Homework Statement

Find f' of ln[((x²+1)^.5) / x(2x³-1)²]

Homework Equations

The Attempt at a Solution

I know the derivative of ln(x) is 1/x... however I don't know how to start this particular problem. I think there is a easier way to solve this without trying to solve the derivative of [((x²+1)^.5) / x(2x³-1)²]..?

You need to use the chain rule. And I don't think there is an "easy" way to do this (i.e. a shortcut). You just need to do a lot of bookkeeping.

Also although you are correct to say that d/dx [ln(x)] = 1/x, that does not really help since the argument of the log function is not 'x' it is a function of x. So we say that if 'u' is a function of x, then d/dx [ln(u)] = 1/u *du/dx.

That is the chain rule applied to the natural log function.

 

[HallsofIvy]

Use the laws of logarithms:

\ln\left[\frac{(x^2+ 1)^{.5}}{x(2x^3-1)^2}\right]= .5 ln(x^2+ 1)- ln(x)- 2ln(2x^3- 1)

That's a little easier to differentiate.

 

[Mark44]

Homework Statement

Find f' of ln[((x²+1)^.5) / x(2x³-1)²]

Homework Equations

The Attempt at a Solution

I know the derivative of ln(x) is 1/x... however I don't know how to start this particular problem. I think there is a easier way to solve this without trying to solve the derivative of [((x²+1)^.5) / x(2x³-1)²]..?

Before differentiating, use the properties of logs so that instead of the log of a quotient, you're working with a difference of logs. You will need to use the chain rule when you actually start differentiating.

 

[HallsofIvy]

Wow, three responses in three minutes.

But, this time, I got in before Mark44!

 

[lude1]

Thanks! :)

 

[Mark44]

Wow, three responses in three minutes.

But, this time, I got in before Mark44!

I'm having a slow day

 

[Saladsamurai]

Wow, three responses in three minutes.

But, this time, I got in before Mark44!

Muh hahaha! But not before me! Now somebody answer my Incomplete Gamma function question!

 

[Susanne217]

Wow, three responses in three minutes.

But, this time, I got in before Mark44!

HallsoftIvy,

If this was the star wars universe you would be yoda and the rest of us younglings!

 

[HallsofIvy]

How did you find out I was short and wrinkly?