Find the derivative of the function:
f(x)=(x^2+9)/square roote of x
f(x)=u/v
f'=u'/v -v'u/v^2
can you do that? u=x^2+9 v=x^.5
u'=2x v'=1/2 x^-.5
Yeah, what I accidently did was I tried to solve it further, but I completely over thought that. Thank you Bob!
To find the derivative of the function f(x) = (x^2 + 9) / √x, we can use the quotient rule of differentiation. The quotient rule states that if we have a function f(x) = g(x) / h(x), where g(x) and h(x) are differentiable functions, then the derivative of f(x) can be computed as follows:
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
In our case, g(x) = x^2 + 9 and h(x) = √x. Let's find the derivative step by step.
Step 1: Find the derivative of g(x).
g'(x) = 2x
Step 2: Find the derivative of h(x).
h'(x) = (1/2) * x^(-1/2) = 1 / (2√x)
Step 3: Apply the quotient rule.
f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x))^2
= (2x * √x - (x^2 + 9) * (1 / (2√x)) / (√x)^2
= (2x√x - (x^2 + 9) / (2√x)) / x
= (2x√x - (x^2 + 9)) / (2x√x)
= (2x√x - x^2 - 9) / (2x√x)
So, the derivative of the function f(x) = (x^2 + 9) / √x is f'(x) = (2x√x - x^2 - 9) / (2x√x).