How to find g'(x) when its given g(x)=6/x^2 + 1/3x + 2?

Answer 1

Since a negative exponent in the numerator is the same thing as a positive exponent in the denominator, let's make it where negative exponents are in numerators to make things easier as it is easier to think this way. So, we have 6x^-2 + (1x^-1)/3

When we go from g(x) from g'(x), we are taking the derivative of g(x) once. When we take the derivative of a term, we multiply the coefficient by the variable's exponent, and then we subtract that variable's exponent by one. So, we will get -12x^-3 - 1x^-2/3 + 0 = -12/x³ - 1/3x²

So, g'(x) = -12/x³ - 1/3x²

Answer 2

Rewrite 6/(x²) as 6x^-2

Rewrite 1/(3x) as (1/3)x^-1

Now us the power rule to take the derivative, for f(x) = xⁿ, f '(x) = nx^n-1