Derivative
Take the deriviative of the following equation:
4x+5
4
Find the equation of the normal to the curve of the function below at the point where x=4
y=x + 6/x
(dy)/ (dx)= 1 -6x^-2
1-6(4)^-2= 5/8
bot = -8/5
y=4
y= 4+6/4 = 5.5
y-5.5= -8/5(x-4)
(-8x)/5 -9/10
Find the value(s) of x for which dy/dx=0.
y= x^2-6x
2x-6=0
x=-6/3= -2
If
y=x^2-3x
dy/dx when x=4
dy/dx =2x-3
2(4)-3 = 5
2+2
4
find f'.
2x^3+4x+ 33
6x^2+4
The normal line means you are looking for what?
A gradient that has a perpendicular slope to that of the original equation.
y=12x-2x^2
12-4X
0=12-4X
12/-4=X
X=-3
If
6x-x^3 +4
find
dy/ dx
when x=0
dy/ dx = 6-3x^2
dy/dx = 6- 3(0)^2 = 6
3-3
0
Find
dy/dx
4/x+2x
-4x^-2+2
If given the following slope of the original equation
y= 3x+5
What would be the normal gradient?
-1/3
2X^3-9X^2+12X-7
6X-18X+12
6(X^2-3X+2)
6(X-2)(X-1)
X=2 AND X=1
W= 7.25p^3
Find
(dw)/(dp)
at p=-2
(dw)/(dp)= 21.75p^2
21.75(-2)^2
87
10+10
20
take the derivative of
2/(3x^2) -7x+33
(-4x^-3)/3-7
If given the following slope of the original equation
y= -(2x)/3 + 8
What would be the normal gradient?
3/2
Y=X(9+3X-X^2)
(-1,-5) MIN (3, 27) MAX
If
y=2x(5x+4)
find the value of
dy/dx
when x=-1
10x^2 +8x
y'= 20x+8
20(-1)+8= -12
2 xx 2/(3*x^2)
4/(3*x^2)
find f' of the following function. Answer must be full simplified if it can be.
1/(2x^2) - 3/(4x)+ 25
-x +3/4x^-2
If given the following slope of the original equation
y=4/7x+10
What would be the normal gradient?
-7/4
identify the max and the min
x^3-9x^2+24x-20
(2, 0) and (4, -4)
V=7r^3-8/r
Find
(dv)/( dr)
at r=2
(dv)/(dr) = 8r-18r^-2
8(3)-18(3)^-2
22
100-25-75
0