y=3 cos(x) y' = -3 sin(x)
D(y)= sin 2x
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
That means you must take the derivative of the derivative. In this case, you must use the product rule. y = 6x sin x y'= 6[x (sin x)' + (x)' sin x] = 6[x cos x + sin x] y'' = 6[x (cos x)' + (x)' cos x + cos x] = 6[x (-sin x) + cos x + cos x] = 6[-x sin x + 2 cos x]
x equals -12 and y equals 1/4 of -12, so y = -3.
Given y = tan x: dy/dx = sec^2 x(secant of x squared)
- 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)
(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.
y' = (sec(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).
D(y)= sin 2x
Derivative of sin x = cos x, so chain rule to derive 8x = 8 , answer is 8cos8x