Question

Could someone help walk me through a derivative. This one is giving me trouble. d/dx (3x^2+6x+6)/sqrtx

Answer 1

, http://tutorial.math.lamar.edu/Classes/CalcI/ProductQuotientRule.aspx#Deriv_PQRule_Ex2a

Answer 2

i do same things... i think this site is good

Answer 3

\[\frac{ d }{ dx }(3x ^{2}+6x+6)/\sqrt{x}\]

Answer 4

\[\sqrt x=x^{1/2}\]and\[\frac{x^a}{x^ b}=x^{a-b}\]use that to simplify each term, then apply the power rule

Answer 5

let f(x)=3x^2+6x+6) and g(x)=\[\sqrt{x}\]=x^1/2

Answer 6

\[{3x^2+6x+6\over\sqrt x}={3x^2+6x+6\over x^{1/2}}=3x^{2-1/2}+6x^{1-1/2}+6x^{0-1/2}\]

Answer 7

\[=3x^{3/2}+6x^{1/2}+6x^{-1/2}\]\[\frac d{dx}(3x^{3/2}+6x^{1/2}+6x^{-1/2})=?\]

Answer 8

g(x)* f'(x) - g'(x)* f(x)/g(x)^2

Answer 9

Got (9sqrtx/2)+(3/sqrtx)-(3/sqrt(x^3)). Thanks, I never knew that x^a/x^b=x^(a-b). But, I guess it follows from 1/x^(-1)=x.