- 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
(xlnx)' = lnx + 1
x=2
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
if y=aex+be2x+ce3x then its derivative dy/dx or y' is: y'=aex+2be2x+3ce3x
sec(x)tan(x)
- 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 = 36cot(x)Y' = dy/dx36cot(x)= - 36csc2(x)==========
Find the derivative of Y and then divide that by the derivative of A
y=3 cos(x) y' = -3 sin(x)
In this case, you'll need to apply the chain rule, first taking the derivative of the tan function, and multiplying by the derivative of 3x: y = tan(3x) ∴ dy/dx = 3sec2(3x)
(xlnx)' = lnx + 1
x=2
Because the derivative of e^x is e^x (the original function back again). This is the only function that has this behavior.
There are several steps involved in how one can solve the derivative x plus y - 1 equals x2 plus y2. The final answer to this math problem is y'(x) = (1-2 x)/(2 y-1).
D(y)= sin 2x
tan(b) = x/sqrt(y^2-x^2)