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=nxn1

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=nxn1
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=nxn1
gives us:
7dxd(x6)=7×6x5=42x5
Note: We can do this in one step:
dxdy=42x5
We can write: dxdy=42x5 OR y=42x5. They mean the same thing.

Example 2

Find the derivative of y = 3x5 − 1
y=3x51
Now,
dxd(3x5)=3×5x4=15x4
And since dxdc=0, we can write:
dxd(1)=0
So
dxdy=dxd(3x51)=15x4

Example 3

Find the derivative of
y=13x46x3x1
Now, taking each term in turn:
dxd(13x4)=52x3 (using dxdxn=nxn1)
dxd(6x3)=18x2 (using dxdxn=nxn1)
dxd(x)=1 (since x=(x1) and so the derivative will be (x0)=1)
dxd(1)=0 (since dxdc=0)
So
dxdy=52x318x21

Example 4

Find the derivative of
y=41x8+21x432
y=41x8+21x432
Differentiating term by term, we have:
dxd(41x8)=48x7=2x7
dxd(21x4)=24x3=2x3
dxd(32)=0 (this is the derivative ofa constant)
So
dxdy=dxd(41x8+21x432)=2x7+2x3

Example 5

Evaluate the derivative of
y=x49x25x
at the point (3,15).
y=x49x25x
So
dxdy=4x318x5
At the point where x=3, the derivative has value:
dxdy=4(3)318(3)5
=4×2718×35
=49
This means that the slope of the curve y=x49x25x at x=3 is 49.

Example 6

Find the derivative of the function
y=x1/4x2
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=x1/4x2
We can write this as:
y=x1/42x1
Differentiating gives us:
dxdy=41x4112(1)x11
=41x3/4+2x2
=4x3/41+x22

Exercise

Find the equation of the tangent to the curve y=3xx3 at x=2.
Now y=3xx3
dxdy=33x2 and the value of this derivative at x=2 is given by:
dxdy=33(2)2=9
Since y=3xx3, then when x=2y=2.
So we need the equation of the line passing through (2,2) with slope 9.
Using the general equation of the line yy1=m(xx1), we have:
y+2=9(x2)
So the required equation is:
y=9x+16
Or, in general form: 9x+y16=0.

 
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