find the first dericative of y=(2500+500(sin100pix))/x
use the quotient rule ....
dy/dx = [ x(0 + 500cos(100πx)(100π)) - (2500 + 500sin(100πx) ) ]/x^2
= ....
expand and simplify it a bit
y=(2500+500(sin100pix))/x
y=2500/x+500(sin100pix))/x
y'=-2500/x^2 +500*100pi*cosx/x -500(sin100pix/x^2)
check that.
https://www.derivative-calculator.net/
To find the first derivative of y = (2500 + 500(sin(100πx))) / x, we can use the quotient rule of differentiation. The quotient rule states that for a function of the form u/v, the derivative is given by:
dy/dx = (v * du/dx - u * dv/dx) / (v^2)
In this case, u = (2500 + 500(sin(100πx))) and v = x.
Let's calculate the derivative step by step:
Step 1: Find the derivative of u with respect to x (du/dx).
Using the chain rule, we differentiate the outer function (500(sin(100πx))) and the inner function (100πx) separately:
du/dx = 500 * d(sin(100πx))/dx
= 500 * cos(100πx) * d(100πx)/dx
= 500 * cos(100πx) * 100π
= 50000πcos(100πx)
Step 2: Find the derivative of v with respect to x (dv/dx).
Since v = x, the derivative dv/dx is 1.
Step 3: Substitute the values into the quotient rule formula:
dy/dx = (v * du/dx - u * dv/dx) / (v^2)
= (x * (50000πcos(100πx)) - (2500 + 500(sin(100πx))) * 1) / (x^2)
= (50000πx * cos(100πx) - (2500 + 500(sin(100πx)))) / (x^2)
Therefore, the first derivative of y = (2500 + 500(sin(100πx))) / x is (50000πx * cos(100πx) - (2500 + 500(sin(100πx)))) / (x^2).