Sunday

February 1, 2015

February 1, 2015

Posted by **Anonymous** on Sunday, January 25, 2009 at 12:57pm.

f(x) = 28x*ln√x + 42

f'(x) = 28*ln(x)^(1/2) + (1/(x^(1/2))*(1/2(x)^(-1/2)28x)

f'(x) = 28*ln(x)^(1/2) + (14x*x^(-1/2))/x^(1/2)

f'(x) = 28*ln(x)^(1/2) + (14x^(-1/2))/x^(1/2)

f'(x) = 28*ln(x)^(1/2) + (14x^(-1)

f'(x) = 28*ln(x)^(1/2) + 1/14x

... how do I simplify further to "14(1 + lnx)"??

- Math - Derivatives -
**Damon**, Sunday, January 25, 2009 at 1:09pm28 x ln [sqrt (x+42)] ?maybe? I need more parentheses to understand.

[28 x/sqrt(x+42) ](1/2)x/sqrt(x+42) +28 ln(sqrt(x+42)]

14 x^2/(x+42) + 28 ln[sqrt(x+42]

- Math - Derivatives -
**Anonymous**, Sunday, January 25, 2009 at 1:27pmThe original equation is...

f(x) = (28x)(ln(sqrt(x)) + 42

- Math - Derivatives -
**Damon**, Sunday, January 25, 2009 at 3:35pmWell then, the 42 is irrelevant for our purposes since its derivative is zero.

so

28 x (d/dx ln x^.5) + 28 ln x^.5

28 x [ (1/x^.5) .5 x^-.5 ] + 28 ln x^.5

28 x (.5/x) + 28 ln x^.5

14 + 28 ln sqrt x

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