Common derivatives :
A. Derivative of a Constant
dxdc=0
This is basic. In English, it means that if a quantity has a constant value, then the rate of change is zero.
Example : a
dxd(6)=0
B. Derivative of n-th power of x
dxdxn=nxn−1
Example : b
dxdx5=5x4
C. Derivative of Constant product
dxd(cy)=cdxd(y)=cdxdy
Here, y is some function of x. It means that if we are finding the derivative of a constant times that function, it is the same as finding the derivative of the function first, then multiplying by the constant.
Example c
If y=x7, then dxdy=dxd(x7)=7x6.
Applying the new rule (c), we have:
dxd(3y)=dxd(3x7)
=3dxd(x7)
=3dxdy
=(3)(7x6)
=21x6
D. Derivative of a sum
dxd(u+v)=dxdu+dxdv
Here, u and v are functions of x. The derivative of the sum is simply equal to the derivative of the first plus derivative of the second. It does not work the same for the derivative of the product of two functions, that we meet in the next section.
Example : d
If u=x2 and v=x9, then:
dxd(u+v)=dxd(x2)+dxd(x9)
=2x+9x8
Derivatives Summary
Constant: | dxdc=0 |
n-th power of x: | dxdxn=nxn−1 |
Constant product: | dxd(cy) =cdxd(y)=cdxdy |
Sum: | dxd(u+v)=dxdu+dxdv |
Further Examples :
Example 1
Find the derivative of y = −7x6
Using the rule
dxd(cy)=cdxd(y)we can take the -7 out the front:
dxd(−7x6)=−7dxd(x6)And
dxdxn=nxn−1gives us:
−7dxd(x6)=−7×6x5=−42x5Note: We can do this in one step:
dxdy=−42x5We can write: dxdy=−42x5 OR y′=−42x5. They mean the same thing.
Example 2
Find the derivative of y = 3x5 − 1
y=3x5−1Now,
dxd(3x5)=3×5x4=15x4And since dxdc=0, we can write:
dxd(−1)=0So
dxdy=dxd(3x5−1)=15x4
Example 3
Find the derivative of
y=13x4−6x3−x−1
Now, taking each term in turn:
dxd(13x4)=52x3 (using dxdxn=nxn−1)
dxd(−6x3)=−18x2 (using dxdxn=nxn−1)
dxd(−x)=−1 (since −x=−(x1) and so the derivative will be −(x0)=−1)
dxd(−1)=0 (since dxdc=0)
So
dxd(13x4)=52x3 (using dxdxn=nxn−1)
dxd(−6x3)=−18x2 (using dxdxn=nxn−1)
dxd(−x)=−1 (since −x=−(x1) and so the derivative will be −(x0)=−1)
dxd(−1)=0 (since dxdc=0)
So
dxdy=52x3−18x2−1
Example 4
Find the derivative of
y=−41x8+21x4−32
y=−41x8+21x4−32Differentiating term by term, we have:
dxd(−41x8)=−48x7=−2x7
dxd(21x4)=24x3=2x3So
dxd(32)=0 (this is the derivative ofa constant)
dxdy=dxd(−41x8+21x4−32)=−2x7+2x3
Example 5
Evaluate the derivative of
y=x4−9x2−5x
at the point (3,15).
y=x4−9x2−5x
So
So
dxdy=4x3−18x−5At the point where x=3, the derivative has value:
dxdy=4(3)3−18(3)−5This means that the slope of the curve y=x4−9x2−5x at x=3 is 49.=4×27−18×3−5=49
Example 6
Find the derivative of the function
y=x1/4−x2
In this case we have fractions and negative numbers for the powers of x. (So it is not a polynomial).
The differentiation rules still apply.
y=x1/4−x2
We can write this as:
y=x1/4−2x−1
Differentiating gives us:
The differentiation rules still apply.
y=x1/4−x2
We can write this as:
y=x1/4−2x−1
Differentiating gives us:
dxdy=41x41−1−2(−1)x−1−1=41x−3/4+2x−2=4x3/41+x22
Exercise
Find the equation of the tangent to the curve y=3x−x3 at x=2.
Now y=3x−x3
dxdy=3−3x2 and the value of this derivative at x=2 is given by:
dxdy=3−3(2)2=−9
Since y=3x−x3, then when x=2, y=−2.
So we need the equation of the line passing through (2,−2) with slope −9.
Using the general equation of the line y−y1=m(x−x1), we have:
dxdy=3−3x2 and the value of this derivative at x=2 is given by:
dxdy=3−3(2)2=−9
Since y=3x−x3, then when x=2, y=−2.
So we need the equation of the line passing through (2,−2) with slope −9.
Using the general equation of the line y−y1=m(x−x1), we have:
y+2=−9(x−2)So the required equation is:
y=−9x+16Or, in general form: 9x+y−16=0.