Math  Derivatives
posted by Anonymous .
Differentiate the given function. Simplify the answer.
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)"??

28 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] 
The original equation is...
f(x) = (28x)(ln(sqrt(x)) + 42 
Well 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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