A linear function, for example y(x) = ax + b has the first derivative a.
There should not be any y in the derivative itself since y or y(x) is the function whose derivative you are finding.
The "double prime", or second derivative of y = 5x, equals zero. The first derivative is 5, a constant. Since the derivative of any constant is zero, the derivative of 5 is zero.
The derivative with respect to 'x' is 4y3 . The derivative with respect to 'y' is 12xy2 .
assuming that you are referring to a function: f(y) = y×y or better put: f(y) = y2 Then it's derivative would be: f'(y) = 2y
y"+y'=0 is a differential equation and mean the first derivative plus the second derivative =0.Look at e-x the first derivative is -e-xThe second derivative will be e-xThe sum will be 0
A linear function, for example y(x) = ax + b has the first derivative a.
Afetr you take the first derivative you take it again Example y = x^2 dy/dx = 2x ( first derivative) d2y/dx2 = 2 ( second derivative)
All it means to take the second derivative is to take the derivative of a function twice. For example, say you start with the function y=x2+2x The first derivative would be 2x+2 But when you take the derivative the first derivative you get the second derivative which would be 2
Derivative of e2x - y = 2e2x - dy/dx
There should not be any y in the derivative itself since y or y(x) is the function whose derivative you are finding.
The "double prime", or second derivative of y = 5x, equals zero. The first derivative is 5, a constant. Since the derivative of any constant is zero, the derivative of 5 is zero.
The first derivative is m and the second is 0 so the third is also 0.
If y = 3x +- 1, the derivative with respect to x is y' = 3.
The derivative with respect to 'x' is 4y3 . The derivative with respect to 'y' is 12xy2 .
assuming that you are referring to a function: f(y) = y×y or better put: f(y) = y2 Then it's derivative would be: f'(y) = 2y
Find the derivative of Y and then divide that by the derivative of A