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Differentiate the function: g(u)=sqrt(2)u+sqrt(3u)
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\[(x^{a})' = ax^{a-1}\] sqrt(x) = x^0.5 does it helps ?
Oh god, I said negative exponents. I meant as a fraction. ^like above.
is that the same rule as d/dx(x^n)=nx^(n-1)? and how'd you get sqrt(x)=x^0.5?
x^(1/2) is the definition of sqrt(x)
And yes differentiating it will be the same rule.
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where'd you get x^1/2 though?
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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