find the derivative of the function y=x^x
Hence fin the coordinates of the turning point of the funcion y-x^x. Give the values correct to two decimal places
The derivative of that function is
y ' = (ln x + 1)*x^x
I will provide a website with the derivation separately.
I don't know what you mean by the "turning point of the function y-x^x". That function would always be zero. Do you mean the minimum of the function y = x^x?
See this website for how to differentiate the function:
http://www.analyzemath.com/calculus/Differentiation/first_derivative.html
To find the derivative of the function y = x^x, we can use logarithmic differentiation. The steps to find the derivative are as follows:
Step 1: Take the natural logarithm of both sides of the equation:
ln(y) = ln(x^x)
Step 2: Use the logarithmic property to simplify the right side of the equation:
ln(y) = x * ln(x)
Step 3: Differentiate both sides of the equation implicitly with respect to x:
(1/y) * dy/dx = ln(x) + 1
Step 4: Solve for dy/dx (the derivative of y with respect to x):
dy/dx = y * (ln(x) + 1)
Since y = x^x, we substitute this back into the equation:
dy/dx = x^x * (ln(x) + 1)
So, the derivative of the function y = x^x is dy/dx = x^x * (ln(x) + 1).
To find the coordinates of the turning point of the function y - x^x, we need to find the critical point(s) by setting the derivative equal to zero:
dy/dx = x^x * (ln(x) + 1) = 0
For x^x to be zero, x must be zero as well. However, x cannot be negative because it causes x^x to be undefined. So, the only possible critical point is x = 0.
To find the y-coordinate of the turning point, substitute the x-coordinate into the original function y = x^x:
y = 0^0 = 1
Hence, the turning point of the function is at (0, 1) with values correct to two decimal places.