Whats the slope of the tangent to the curve y=7/(square root of x) at the point where x=a>0?
To find the slope of the tangent to the curve at a specific point, you need to determine the derivative of the function and evaluate it at that point. Let's start by finding the derivative of the given function y = 7/√x using the quotient rule.
The quotient rule states that the derivative of a quotient of two functions, u(x) and v(x), is given by:
(dy/dx) = (v(x) * du/dx - u(x) * dv/dx) / (v(x))^2.
In this case, u(x) = 7 and v(x) = √x. Applying the quotient rule, we obtain:
(dy/dx) = (√x * 0 - 7 * (1/2√x)) / (√x)^2
Simplifying the expression, we have:
(dy/dx) = -7 / (2x√x) = -7/(2x^(3/2))
Now that we have the derivative of the function, we can evaluate it at the given point x = a. Substituting x = a into the derivative expression, we get:
m = (dy/dx) evaluated at x = a = -7/(2a^(3/2))
Therefore, the slope of the tangent to the curve y = 7/√x at the point where x = a is -7/(2a^(3/2)).