For any function, With utilization of the product rule for functions of one variable (eq1), show / prove how it could be viewed as the result of the chain rule applied to a particular function of two variables(eq2). Eq 1 notation: The Product Rule If f and g are both differentiable, then (fg)' = fg' + gf' d d Sg(x}] = f(x) [g(x)] + g(x) [S(x)] dx Eled 2 The Chain Rule (Case 1) Suppose that z = f(x, y) is a differentiable func- tion of x and y, where x = g(t) and y = h(t) are both differentiable functions of t. Then z is a differentiable function of t and of dx ax dt dz af dy dt ду di

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
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For any function , With utilization of the product rule for functions of one
variable (eq1), show / prove how it could be viewed as the result of the
chain rule applied to a particular function of two variables(eq2).
Eq 1
notation:
The Product Rule If f and g are both differentiable, then
(fg)' = fg' + gf'
d
d
d
[S(x)g(x)] = f(x) – [g(x)] + g(x) [f(x)]
%3D
dx
dx
dx
Eed 2 The Chain Rule (Case 1) Suppose that z = f(x, y) is a differentiable func-
tion of x and y, where x = g(t) and y = h(t) are both differentiable functions of t.
Then z is a differentiable function of t and
dz
af dx
of dy
+
dt
dx dt
ду di
Transcribed Image Text:For any function , With utilization of the product rule for functions of one variable (eq1), show / prove how it could be viewed as the result of the chain rule applied to a particular function of two variables(eq2). Eq 1 notation: The Product Rule If f and g are both differentiable, then (fg)' = fg' + gf' d d d [S(x)g(x)] = f(x) – [g(x)] + g(x) [f(x)] %3D dx dx dx Eed 2 The Chain Rule (Case 1) Suppose that z = f(x, y) is a differentiable func- tion of x and y, where x = g(t) and y = h(t) are both differentiable functions of t. Then z is a differentiable function of t and dz af dx of dy + dt dx dt ду di
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