G( x)=sqrt(3x) find g'(x)

1 answer:

4 0

Well, g'(x) is just a fancy way of saying "the slope of". In calculus, it's also called "the derivative".
So our original function is
sqrt3x (I'm leaving out the G(x))
First, we'll need to rewrite this in exponential form:
(3x)^1/2
Now, to take a derivative, move the exponent to the front of the expression:
1/2(3x)^1/2
And add 1 to the exponent:
g'(x)=1/2(3x)^3/2
g'(x)=1/2(sqrt27)x^3/2
g'(x)=((sqrt27)/2)x^3/2
I hope it didn't get confusing with all the parentheses! Sorry if it did!
Hope this helps!

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