What is the derivative of square root of 7x

Okay so i know that the square root equals .5 this is how i did it

i brought the .5 to the front of the seven which gave me 7/2 (or 3.5) then i did the power rule n-1 which gave me -1/2 or (-0.5) which would make the square root in the denominator

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My answer is not correct! i get 7/(2square root x)

*** what am i doing wrong???

diff(sqrt(7*x), x) = (1/2)*sqrt(7)/sqrt(x)

The answer is 7/2√(7x)

You left out the 7 under the radical.

Worth remembering that if y = √x then dy/dx = 1/(2√x)

So for your problem use that with the chain rule, differentiating the 7x to get the 7 on top.

If you want to always derive it with fraciotnal powers, then f(x) = (7x)^½, so

f '(x) = ½ (7x)^-½•7 = 7/(2√(7x)) like I first said.

You forgot to use chain rule
Because it is √(7x) INCLUDING 7 so it is the same as
(7x)^(1/2) take derivative
(1/2)(7x)^(-1/2) TIMES ANOTHER 7 for INSIDE
you get
49/(2√x)

Derivative of square root of 7x is:
1/2 square root of 7x

y = √7x = (7x)^(1/2)
y' = f'(x) = (1/2)(7x)^(-1/2)
f'(x) = 1 / (2)(√7x)