Calcs 6
Calcs 6
Calcs 6
Calculus I
Dr Paul May
Chain Rule
dy
du
and
. We then use:
du
dx
dy dy du
dx du dx
Examples
1.
y = (2 x3)4
let u = 2 x3,
dy
= 4u3
du
So
dy
= (4u3).( -3x2)
dx
so that y = u4
and
du
= -3x2
dx
= -12x2 (2 x3)3
y(x) =
1
, i.e. y = (1 x2)1
(1 x 2 )
33
Chemistry 1S
Calculus I
du
= -2x
dx
(ii)
Dr Paul May
1
dy
=- 2
du
u
1
dy
= ( -2x)( dx
1 x2
1 x
)=+
2 2
2x
(1 x 2 ) 2
With this method, we start with the outermost function, and differentiate our way to the
centre, multiplying everything together along the way.
Examples
1.
y = (2 x3)4
think of this as
y = (expression)4
differentiating,
dy
= 4(expression)3
dx
We now look at the expression in the brackets and differentiate that (= -3x2) and multiply
it to our previous answer to give
dy
= 4(2 x3)3 ( - 3x2)
dx
2.
y=
1
(1 x 2 )
= (1 x2)-1
dy
= -(1 x2)-2 ( -2x)
dx
differential of
(...)-1
3.
y=
x2 1
differential of
1 x2
= (x2 - 1)
dy
= (x2 - 1) - 2x
dx
differential of
(...)
differential of
x2 - 1
34
Chemistry 1S
4.
Calculus I
Dr Paul May
Refer back to this later, after we've covered sin and ln.
y = sin{ln(3x2 + 2)}
1
dy
= cos{ln(3x2 + 2)}
6x
2
dx
3x 2
differential of
sin{...}
differential of
ln(...)
differential of
3x2 + 2
5. Exponential Functions
The general expression for an exponential function is
k & a = constants
f(x) = kax
a>1
y
35
35
30
30
25
25
20
20
15
15
10
10
a<1
An example is y = 3x
x0
y1
1
3
2 3 4 ...
9 27 81 ...
One of the most important properties of an exponential function is that the slope of the
function at any value is proportional to the value of the function itself.
In other words,
dy
y(x),
dx
or
dy
= constant y(x)
dx
35
Chemistry 1S
Calculus I
Dr Paul May
Numerical examples
y = 2x, plot the graph and measure the slopes at different values of x.
1.
slope at x
measured
from graph
y
35
x
0
1
2
3
4
30
25
20
15
10
5
0
y
1
slopey
).
dy
= 0.69 y(x)
dx
i.e.
Try it again, but for y = 3x
2.
x
0
1
2
3
4
y
100
80
60
40
y
1
slope
slopey
20
0
dy
= y(x)
dx
The value of a that gives this result is known as e and has the value:
36
Chemistry 1S
Calculus I
Dr Paul May
e = 2.718...
an irrational number.
e is actually calculated from the following progression formula (see the Algebra part of
the course later if you don understand this equation yet):
e=
n ! 1 1 2 6 24 ....
n0
ex = 1
e x +
x=0
x +
x -
e x 0
x -
x
ex
e- x
0
1
1
1
2.72
0.37
2
7.39
0.14
3
20.1
0.05
4
54.6
0.02
e -x = 1
e -x 0
e -x +
5
148
0.007
-x
y=e
y=e
0
-4
-3
-2
-1
0
-3
-2
-1
37