what is D.E of dy/dx=x/(x^2+9)^1/2 given y=5 n x=4
Answers
Answer:
The Chain Rule
Our goal is to differentiate functions such as
y = (3x + 1)10
The Chain Rule
If
y = y(u)
is a function of u, and
u = u(x)
is a function of x then
dy dy du
=
dx du dx
In our example we have
y = u10
and
u = 3x + 1
so that
dy/dx = (dy/du)(du/dx)
= (10u9) (3) = 30u9 = 30 (3x+1)9
Proof of the Chain Rule
Recall an alternate definition of the derivative:
Examples
Find f '(x) if
f(x) = (x3 - x + 1)20
f(x) = (x4 - 3x3 + x)5
f(x) = (1 - x)9 (1-x2)4
(x3 + 4x - 3)7
f(x) =
(2x - 1)3
Solution:
Here
f(u) = u20
and
u(x) = x3 - x + 1
So that the derivative is
[20u19] [3x2 - 1] = [20(x3 - x + 1)19] [3x2 - 1]
Here
f(u) = u5
and
u(x) = x4 - 3x3 + x
So that the derivative is
[5u4] [4x3 - 9x2 + 1] = [5(x4 - 3x3 + x)4] [4x3 - 9x2 + 1]
Here we need both the product and the chain rule.
f'(x) = [(1 - x)9] [(1 - x2)4]' + [(1 - x)9] ' [(1 - x2)4]
We first compute
[(1 - x2)4] ' = [4(1 - x2)3] [-2x]
and
[(1 - x)9] ' = [9(1 - x)8] [-1]
Putting this all together gives
f'(x) = [(1 - x)9] [4(1 - x2)3] [-2x] - [9(1 - x)8] [(1 - x2)4]
Here we need both the quotient and the chain rule.
(2x - 1)3 [(x3 + 4x - 3)7] ' - (x3 + 4x - 3)7 [(2x - 1)3] '
f '(x) =
(2x - 1)6
We first compute
[(x3 + 4x - 3)7] ' = [7(x3 + 4x - 3)6] [3x2 + 4]
and
[(2x - 1)3] ' = [3(2x - 1)2] [2]
Putting this all together gives
7(2x - 1)3 (x3 + 4x - 3)6 (3x2 + 4) + 6(x3 + 4x - 3)7 (2x - 1)2
f '(x) =
(2x - 1)6
Step-by-step explanation: