Find the slope m of the tangent to the curve

y = 5/ sqrt (x) at the point where
x = a > 0.
what is the slope?

y = 5x^(-1/2)

dy/dx = (-5/2)x^(-3/2)

so when x=a

slope = (-5/2)x^(-3/2)

-5/2a^3/2

-5/(a)(2 sqrt a)

WRONG

Your dumb @phillip

To find the slope (m) of the tangent to the curve y = 5/√(x) at the point where x = a, we can differentiate the given function with respect to x.

1. Start with the given equation: y = 5/√(x).

2. To find the derivative (dy/dx), apply the power rule and chain rule.

- Power rule: d/dx (√(x)) = (1/2) * x^(-1/2)
- Chain rule: d/dx (5/√(x)) = (5) * (1/2) * x^(-1/2) = (5/2) * x^(-1/2)

3. Simplify the derivative expression to get: dy/dx = (5/2) * x^(-1/2).

4. Since we need to find the slope at the point where x = a, substitute a into the derivative expression: dy/dx = (5/2) * a^(-1/2).

5. This expression represents the slope (m) of the tangent at the given point where x = a. Therefore, the slope (m) is (5/2) * a^(-1/2).

the answer is -5/a^2