Difference Quotient and Derivative of a Function

Question

Difference Quotient and Derivative of a Function

Answer 1

1 y=x-2x^(-1/2)

y' = 1 +x^(-3/2)

2 g' = sqr2 +sqr3/2u^(1/2)

3 g' = 2/3t^(1/3) -1/6t^(5/6)

convert each fraction into two numbers, then take the derivative, it's easier that way than dealing with the fractions

the difference quotient gets even messier if you don't remove the fractions first

y = (x^2 - 2sqrx)/x = x^2/x -2sqx/x = x -2x^(-1/2)

y'= 1 +1/x^(-3/2)

y(x) = (x^2 - 2sqrx)/x

y(x+h) = ((x+h)^2 - 2sqr(x+h))/(x+h)

y(x+h) - y(x) = (x^2 +2xh + h^2 - 2sqr(x+h))/(x+h) - (x^2 -2sqrx)/x

then divide all that by h to get the difference quotient

but if you simplify first it's still very messy