Find the derivative of Y and then divide that by the derivative of A
- the derivative with respect to x is 40y - The derivative with respect to Y is 40xSo, since both x and y equal 2, both derivatives yield 40*2 = 80
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
(xlnx)' = lnx + 1
if y=aex+be2x+ce3x then its derivative dy/dx or y' is: y'=aex+2be2x+3ce3x
y=(2x)2 y=2(2x) y=4x
Because the derivative of e^x is e^x (the original function back again). This is the only function that has this behavior.
y=3 cos(x) y' = -3 sin(x)
D(y)= sin 2x
sec(x)tan(x)
Yes, but only if the domain is the real numbers. The derivative is y = 1.
y is a sum of constants and so is itself a constant. Its derivative is, therefore, zero.