6.
Dierentiation
6.1.
Denition
Given a function f (x), its graph is the curve y = f (x).
The tangent at a point on the curve, is the line through the point which touches the curve.
The gradient of the curve at a point is the gradient of the tangent.
The derivative of y is the function which gives the gradient of the curve at the point with
a given x-coordinate.
d
dy
Notation: , or y 0 , or (f (x)) or f 0 (x)
dx
dx
6.1. y = x .
If P is a point on the curve with x-coordinate x, then P is (i.e., has coordinates) (x, x3 ).
A nearby point Q has x-coordinate
x + δx where δx is a small number. Its coordinates are
3
therefore x + δx, (x + δx) .
Moving from P to Q, the change (`increment') in x is δx; the corresponding change in y is
3
Example
δy = (x + δx)3 − x3 = 3x2 δx + 3x δx2 + δx3 .
Therefore the line P Q has gradient
δy
3x2 δx + 3xδx2 + δx3
=
= 3x2 + 3xδx + δx2
δx
δx
Letting δx tend to 0, this gradient tends to 3x2 , so that
δy
dy
= lim
= 3x2 .
dx δx→0 δx
For example the point (2, 8) is on the curve. The gradient of the curve at this point is
3 × 22 = 12.
6.2.
Basic rules
You should know already:
dy
= 0.
dx
dy
2. For any number n, if y = xn then
= nxn−1 ; this is true for any n, positive or
dx
1. If y =constant, then
negative, integer or not an integer.
dy
du dv
=
± .
3. If y = u ± v , then
dx
dx
dx
dy
du
4. If y = au with a constant, then
=a .
dx
dx
Examples.
dy
If y = x5 then
= 5x4 .
dx
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√
√
dy
x = x1/2 , so 1/ x = x−1/2 . Hence, if y = 1/ x then
= (− 12 )x−3/2 .
dx
dy
= 2x + 3x2 .
= x2 + x3 then
dx
dy
2
= 10x then
= 20x.
dx
dy
= 5x then
= 5.
dx
dy
= 10x2 + 5x + 6 then
= 20x + 5.
dx
Note that
If y
If y
If y
If y
6.3.
√
Derivatives of sin and cos
You must use radians when calculating derivatives and integrals.
dy
= cos x.
dx
dy
If y = cos x, then
= − sin x.
dx
If y = sin x, then
Look at their graphs to see if you believe this!
Proof for y = sin x.
The formula for the derivative is
dy
δy
= lim
dx δx→0 δx
Now
δy
sin(x + δx) − sin x
=
δx
δx
sin x cos δx + cos x sin δx − sin x
=
δx
From the picture showing arc length δx and sin δx, we see that,
if δx very small, then sin δx is approximately equal to δx.
Now the formula sin2 δx + cos2 δx = 1 gives
p
√
cos δx = 1 − sin2 δx ∼ 1 − δx2
1
∼ 1 + ( )(−δx2 ) + . . .
2
by the binomial series, so
Therefore
sin x (1 − 12 δx2 ) + (cos x)δx − sin x
δy
∼
δx
δx
1
= cos x − δx
2
dy
= cos x.
dx
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6.4.
The product rule
dy
du
dv
=
v+u .
dx
dx
dx
If y = uv , then
Suppose x is increased by a small amount δx. Then u will increase by a small
amount δu, v by a small amount δv , and y by δy . These are related by
Proof.
y + δy = (u + δu)(v + δv) = uv + δu v + u δv + δu δv
Recall that y = uv , so
δy = δu v + u δv + δu δv
Dividing by δx we get
δu
δv δu δv
δy
=
v+u +
.
δx
δx
δx
δx
δy
dy δu
du
δv
dv
Now letting δx tend to zero,
becomes ,
becomes , and
becomes .
δx
dx δx
dx
δx
dx
dv
δv
which tends to 0. Therefore,
The last term δu tends to δu
δx
dx
dy
du
dv
=
v+u .
dx
dx
dx
Example
6.5.
6.2. Find the derivative of x2 sin x.
The quotient rule
u
v
If y = , then
dy
=
dx
du
v
dx
dv
− u dx
.
v2
Proof.
y + δy =
and y = u/v , so
δy =
Therefore
u + δu
v + δv
u + δu u
(u + δu)v − u(v + δv)
− =
v + δv
v
v(v + δv)
δu v − u δv
.
=
v(v + δv)
Letting δx tend to zero this gives
δu
δv
v − u δx
δy
δx
=
.
δx
v(v + δv)
dy
=
dx
du
v
dx
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dv
− u dx
.
v2
Using the quotient rule we can dierentiate the other trigonometric functions.
dy
= sec2 x.
If y = tan x, then
dx
dy
If y = cosec x, then
= − cot x cosec x.
dx
6.6.
The chain rule
du dy
dy
=
.
dx
dx du
Proof. Suppose x increases by a small amount δx. Then u and y will increase by a small
amounts δu and δy , respectively.
If y is a function of u, where u is a function of x, then
Clearly
δy
δu δy
=
.
δx
δx δu
On letting δx tend to zero, this gives the required formula.
Example
6.3. Suppose y = sin 2x. We write y = sin u where u = 2x. Then
du dy
dy
=
= 2 cos u = 2 cos 2x.
dx
dx du
√
√
Example 6.4. Suppose y =
cos x. We write y = u, where u = cos x.
dy
1
1
Then y = u1/2 , so
= u−1/2 = √ .
du
2
2 u
du
Also u = cos x, so
= − sin x.
dx
Then
dy
du dy
1
=
= − sin x √
dx
dx du
2 u
sin x
=− √
2 cos x
Alternative notation. Suppose f (x) and g(x) are functions. Then we can form the
posite function f (g(x)). Another way to write the chain rule is that
d
f (g(x)) = g 0 (x)f 0 (g(x))
dx
Writing u = g(x) and y = f (u) we see that
dy
d
=
(f (g(x))),
dx
dx
du
= g 0 (x)
dx
and
dy
= f 0 (u) = f 0 (g(x)),
du
so this agrees with the previous form of the chain rule above.
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com-
Example
6.5.
d
(tan(sin x)) = cos x sec2 (sin x)
dx
Special case.
d
(f (ax + b)) = af 0 (ax + b)
dx
Proof: here g is the function with g(x) = ax + b, so g 0 (x) = a.
6.7.
Worked examples
6.6. Find the derivatives of the following functions:
y = (2x + 1)10 ;
y = sin(x + 5);
y = 7x4 ;
y = x3 + 3 cos x + 4;
y = (3x4 + 1) sin x;
y = (1 + cos x)/(x2 + 1);
y = sin(1/x);
√
y = x5 x;
√
y = cos( x);
y = tan2 x;
√
y = x 1 + x2 ;
y = sin4 (3x);
√
y = sin( sin x).
Example
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