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Mathematics 18 Online
OpenStudy (anonymous):

Differentiate the function: g(u)=sqrt(2)u+sqrt(3u)

OpenStudy (anonymous):

\[(x^{a})' = ax^{a-1}\] sqrt(x) = x^0.5 does it helps ?

OpenStudy (cgreenwade2000):

Oh god, I said negative exponents. I meant as a fraction. ^like above.

OpenStudy (anonymous):

is that the same rule as d/dx(x^n)=nx^(n-1)? and how'd you get sqrt(x)=x^0.5?

OpenStudy (cgreenwade2000):

x^(1/2) is the definition of sqrt(x)

OpenStudy (cgreenwade2000):

And yes differentiating it will be the same rule.

OpenStudy (anonymous):

where'd you get x^1/2 though?

OpenStudy (cgreenwade2000):

x^2 is "x squared" x^(1/2) is "square root of x" Just like x^1/3 is the cube root

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