y= (x^3 - 2sqrx)/(x)
y'=
rewrite the square root of x as x^1/2
This is the quotient rule.
demoninator times the derivative of the numerator minus the numerator times the derivative of the denominator all over the denominator squared.
So... We know the derivative will have an x^2 in the denomoniator.
Can you do the top part of the fraction?
who needs the quotient rule?
y = x^2 - 2x^-1/2
y' = 2x + x^-3/2
To find the derivative of y with respect to x (denoted as y'), we can use the quotient rule.
The quotient rule states that if we have a function of the form f(x) = g(x) / h(x), where g(x) and h(x) are both differentiable functions, then the derivative of f(x) is given by:
f'(x) = [g'(x) * h(x) - g(x) * h'(x)] / [h(x)]^2
In our case, g(x) = x^3 - 2sqrt(x) and h(x) = x.
Now, let's find the derivatives of g(x) and h(x) individually:
g'(x) = d/dx(x^3 - 2sqrt(x))
= 3x^2 - 2 * (1/2) * x^(-1/2) (using the power rule and chain rule)
= 3x^2 - x^(-1/2)
h'(x) = d/dx(x)
= 1 (using the power rule)
Now, we can substitute these derivatives into the quotient rule formula to find y':
y' = [g'(x) * h(x) - g(x) * h'(x)] / [h(x)]^2
= [(3x^2 - x^(-1/2)) * x - (x^3 - 2sqrt(x)) * 1] / x^2
= [3x^3 - x^(3/2) - x^3 + 2sqrt(x)] / x^2
= (2sqrt(x) - x^(3/2)) / x^2
Therefore, the derivative of y with respect to x is y' = (2sqrt(x) - x^(3/2)) / x^2.