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# f(x) = (x+5)*lnx  find f'(x)

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f(x) = (x+5)*lnx
find f'(x)

Posted by amal-m on August 17, 2010 at 8:50 AM via web and tagged with equastion, math

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f(x) = (x+5)*lnx

Find f'(x)

First let us assume that:

f(x) = u*v   such that:

u= x+5  ==> u' = 1

v = ln x ==> v' = 1/x

By definition:

f'(x) = u'v + uv'

= 1*lnx + (x+5)*(1/x)

= ln x + 1 + 5/x

==> f'(x) = ln x + 5/x  + 1

Posted by hala718 on August 17, 2010 at 8:53 AM (Answer #1)

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f(x) = (x+5)lnx.

To find f'(x).

Solution:

Here f(x)  in the form of product  of two functions like u(x)v(x) .

So we use {u(x)v(x)}' = u(x) v'(x) +u'(x)v(x).

u(x) =(x+5) , So u'(x) = (x+5) = 1

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

So f(x) = (x+5)lnx = (x+5){lnx)' +(x+5)' lnx

=(x+5)/x +  (1)lnx

=1+5/x +lnx

Posted by neela on August 17, 2010 at 10:41 AM (Answer #2)

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We could remove the brackets and after that, we'll differentiate the function f(x).

f(x) = x*ln x + 5*ln x

For the second term, we'll use the product property of the logarithms:

5*ln x = ln (x^5)

Now, we'll calculate the first derivative of the function:

f'(x) = (x*ln x)' + [ ln (x^5)]'

f'(x) = x'*lnx + x*(ln x)' + (x^5)'/x^5

f'(x) = ln x + x/x + 5x^4/x^5

We'll cancel like terms:

f'(x) = ln x + 1 + 5/x

Posted by giorgiana1976 on August 17, 2010 at 9:35 PM (Answer #3)

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f(x) = (x+5)*lnx

f'(x)=(x+5)1/x + lnx * 1

f'(x)=1+ 5/x + ln x

Posted by jeyaram on August 18, 2010 at 7:00 PM (Answer #4)