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Pre-Calculus and Calculus

derivitives using the quotient rule

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markmil2002
Average Member

USA
11 Posts

Posted - 09/17/2007 : 19:19:19

I'm trying to find the derivitive of f(x) = (x

+ 1)/(ln x)

After doing some math I got -2x/x but I don't think that is correct.

sahsjing
Advanced Member

USA
2399 Posts

Posted - 09/17/2007 : 20:27:52

You should have 1/(lnx)

as a factor. Try it again?

Subhotosh Khan
Advanced Member

USA
9115 Posts

Posted - 09/18/2007 : 09:26:27

quote:
Originally posted by markmil2002

I'm trying to find the derivitive of f(x) = (x + 1)/(ln x)

After doing some math I got -2x/x but I don't think that is correct.

f(x) = u(x)/v(x)

df/dx = (u'*v - v'*u)/v².................(1)

In your case:

u(x) = x² + 1 .................... u' = 2x

v(x) = ln(x) ................................ v' = 1/x

Using these in (1)

df/dx = [(2x) * ln(x) - (x² + 1) * (1/x)]/{ln(x)}²

Now simplify and continue...

markmil2002
Average Member

USA
11 Posts

Posted - 09/18/2007 : 22:12:33

I've made it that far, but I dont see how to simplify. Obviously e will probably be involved to cancel ln, but I cant figure out how to apply it.

quote:
Originally posted by Subhotosh Khan

quote:
Originally posted by markmil2002

I'm trying to find the derivitive of f(x) = (x + 1)/(ln x)

After doing some math I got -2x/x but I don't think that is correct.

f(x) = u(x)/v(x)

df/dx = (u'*v - v'*u)/v².................(1)

In your case:

u(x) = x² + 1 .................... u' = 2x

v(x) = ln(x) ................................ v' = 1/x

Using these in (1)

df/dx = [(2x) * ln(x) - (x² + 1) * (1/x)]/{ln(x)}²

Now simplify and continue...