y=5-2x
y'(u-v)=du/dx-dv/dx
y'(5-2x)=d/dx(5)-d/dx(2x)
-The derivative of 5 is 0 because it's a constant.
-The derivative of 2x is:
d/dx(cu)=c*du/dx where c is a constant.
d/dx(2x)=2*d/dx(x)
y'(5-2x)=(0)-[2*d/dx(x)]
-The derivative of x is:
d/dx(xn)=nxn-1
d/dx(x)=1*x1-1
d/dx(x)=1*x0
d/dx(x)=1*(1)
d/dx(x)=1
y'(5-2x)=-[2*(1)]
y'(5-2x)=-(2)
y'(5-2x)= -2
Derivative of sin x = cos x, so chain rule to derive 8x = 8 , answer is 8cos8x
Because the derivative of e^x is e^x (the original function back again). This is the only function that has this behavior.
Calculate the derivative of the function.Use the derivative to calculate the slope at the specified point.Calculate the y-coordinate for the point.Use the formula for a line that has a specified slope and passes through a specified point.
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If y = 3x +- 1, the derivative with respect to x is y' = 3.
Find the derivative of Y and then divide that by the derivative of A
y=3 cos(x) y' = -3 sin(x)
y is a sum of constants and so is itself a constant. Its derivative is, therefore, zero.
Derivative of sin x = cos x, so chain rule to derive 8x = 8 , answer is 8cos8x
x - y = 52x + y = 7 (use the substitution method)y = x - 52x + y = 7 (substitute x - 5 for y into the second equation)2x + y = 72x + x - 5 = 73x = 12x = 4Thus x = 4 satisfies both the equations of the system.
- 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
y=(2x)2 y=2(2x) y=4x
if y=aex+be2x+ce3x then its derivative dy/dx or y' is: y'=aex+2be2x+3ce3x
Because the derivative of e^x is e^x (the original function back again). This is the only function that has this behavior.
Y = 36cot(x)Y' = dy/dx36cot(x)= - 36csc2(x)==========