# Calculate f'(x) if f(x)=ln[(5x+7)/(5x+3)] x>0 .

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f(x) = ln[5x+7)/(5x+3)]

We know that ln a/b = ln a - ln b

==> f(x) = ln (5x+7) - ln (5x+3)

==> f'(x) = 5/(5x+7) - 5/(5x+3)

= 5[1/(5x+7) - 1/(5x+3)]

= 5[(5x+3)-*5x+7)/(5x+7)(5x+3)

= 5(-4)/(5x+7)(5x+3)

= -20/(5x+7)(5x+3)

Here we use the relations ln(a/b)=lna-lnb and (d ln(ax+b)/dx)=(d(ax+b)/dx)/(ax+b)

f(x)=ln{(5x+7)/(5x+3)}= ln(5x+7)-ln(5x+3)

f'(x)=f'{ln(5x+7)}-f'{ln(5x+3)}

=5/(5x+7)-5/(5x+3)

=5*(5x+3-5x-7)/(5x+7)(5x+3)

=(-20)/(5x+7)(5x+3)

Because f(x) contains the logarithm of a quotient, we'll apply the quotient rule:

ln(f/g)=ln f -ln g.

ln[(5x+7)/(5x+3)] = ln(5x+7) - ln (5x+3)

f'(x) = [ln(5x+7) - ln (5x+3)]'

f'(x) = [ln(5x+7)]' - [ln (5x+3)]'

f'(x)= (5x+7)'/(5x+7) - (5x+3)'/(5x+3)

f'(x) = 5/(5x+7) - 5/(5x+3)

We'll factorize:

f'(x) = 5*[1/(5x+7) - 1/(5x+3)]

f'(x) = 5(5x+3-5x-7)/(5x+7)(5x+3)

**f'(x) = -20/(5x+7)(5x+3)**

f(x) = ln[(5x+7)/(5x+3)] , x> 0.

To find f'(x).

Solution:

f(x) = ln[(5x+7)/5x+3)] = ln(5x+7) - ln (5x+3).

f '(x) = {ln ln(5x+7) - ln (5x+3).} = { ln(5x+7)}' - {ln (5x+3)}'

= 1* (5x+7)'/(5x+7) - 1*(5x+3)/5x+3), as ln(nx+a) = (nx+a)'/(nx+a), n is a number not zero.

= 5/(5x+7) - 5/(5x+3)

= 5 (5x+3-(5x+7))/((5x+7(5x+3))}

= 5(-2)/(5x+7)(5x+3)

=

5(-2)/(5x+7)(5x+3)

= -10/(5x+7)(5x+3)