the derivative of y= x^pie
Using the power rule of differentiation, we can find the derivative of y = x^(π):
dy/dx = πx^(π-1)
if point (a,f(a)) is a critical value of the function y=f(x), and f^2(a)>0, then the point is said to be a local
minimum value of the function.
Explanation: A critical value of a function is a point where the derivative of the function is equal to zero or undefined. It can either be a local minimum, maximum, or a point of inflection.
If f^2(a) > 0, it means that the second derivative of the function at point 'a' is positive, which indicates that the function is "concave up" at this point.
When the function is concave up and the slope of the tangent line is zero at a critical point, the critical point is a local minimum value of the function. This is because the function is increasing on either side of the critical point, and the minimum occurs at the point where the slope of the function changes from negative to positive.
the vertical asymptote of the curve y = x/(x-5) is
The vertical asymptote of the curve given by the equation y = x/(x-5) is x = 5.
Explanation:
A vertical asymptote is a vertical line that the curve approaches as the values of x get larger or smaller without bound.
In this case, the function has a vertical asymptote wherever the denominator (x-5) is equal to zero, because division by zero is undefined. Solving for x:
x - 5 = 0
x = 5
Therefore, the function has a vertical asymptote at x = 5. As x approaches 5 from either side, the function approaches infinity or negative infinity depending on the direction of approach.
To find the derivative of y = x^π, we can use the chain rule. The chain rule states that if we have a composite function, such as y = f(g(x)), then the derivative of y with respect to x is given by the derivative of the outer function f' evaluated at the inner function g(x), multiplied by the derivative of the inner function g'(x).
In this case, y = x^π can be written as y = (x^π) = (e^(π ln(x))). Now we can treat it as a composite function with the outer function f(u) = e^u and the inner function g(x) = π ln(x).
To find the derivative of the outer function f(u) = e^u, we know that the derivative of e^u is simply e^u.
To find the derivative of the inner function g(x) = π ln(x), we can use the chain rule once again. The derivative of ln(x) is 1/x, and the derivative of π ln(x) is π/x.
Now, applying the chain rule, we have:
dy/dx = f'(g(x)) * g'(x)
= e^(π ln(x)) * π/x
= π e^(π ln(x))/x
Thus, the derivative of y = x^π is dy/dx = π e^(π ln(x))/x.