In calculus, the chain rule is a formula to compute the derivative of a composite function. That is, if f and g are differentiable functions, then the chain rule expresses the derivative of their composite f ∘ g — the function which maps x to
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{\displaystyle f(g(x))}
— in terms of the derivatives of f and g and the product of functions as follows:
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{\displaystyle (f\circ g)'=(f'\circ g)\cdot g'.}
Alternatively, by letting h = f ∘ g (equiv., h(x) = f(g(x)) for all x), one can also write the chain rule in Lagrange's notation, as follows:
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{\displaystyle h'(x)=f'(g(x))g'(x).}
The chain rule may also be rewritten in Leibniz's notation in the following way. If a variable z depends on the variable y, which itself depends on the variable x (i.e., y and z are dependent variables), then z, via the intermediate variable of y, depends on x as well. In which case, the chain rule states that:
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{\displaystyle {\frac {dz}{dx}}={\frac {dz}{dy}}\cdot {\frac {dy}{dx}}.}
More precisely, to indicate the point each derivative is evaluated at,
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{\displaystyle \left.{\frac {dz}{dx}}\right_{x}=\left.{\frac {dz}{dy}}\right_{y(x)}\cdot \left.{\frac {dy}{dx}}\right_{x}}
.
The versions of the chain rule in the Lagrange and the Leibniz notation are equivalent, in the sense that if
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{\displaystyle z=f(y)}
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{\displaystyle y=g(x)}
, so that
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{\displaystyle z=f(g(x))=(f\circ g)(x)}
, then
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{\displaystyle \left.{\frac {dz}{dx}}\right_{x}=(f\circ g)'(x)}
and
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{\displaystyle \left.{\frac {dz}{dy}}\right_{y(x)}\cdot \left.{\frac {dy}{dx}}\right_{x}=f'(y(x))g'(x)=f'(g(x))g'(x).}
Intuitively, the chain rule states that knowing the instantaneous rate of change of z relative to y and that of y relative to x allows one to calculate the instantaneous rate of change of z relative to x. As put by George F. Simmons: "if a car travels twice as fast as a bicycle and the bicycle is four times as fast as a walking man, then the car travels 2 × 4 = 8 times as fast as the man."In integration, the counterpart to the chain rule is the substitution rule.
Hi everyone
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Homework Statement
I am facing problem in applying the chain rule.
The question which I am trying to solve is,
" Find the second derivative of "
Homework Equations
The Attempt at a Solution
So, differentiated it the first time,
[BY CHAIN RULE]
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1. The problem statement, all variables, and given/known data
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Homework Equations
Chain rule $$\frac{dz}{dx} = \frac{dz}{dy}...
Homework Statement
Homework Equations
The Attempt at a Solution
I am trying to repair my rusty calculus. I don't see how du = dx*dt/dt, I know its chain rule, but I got (du/dx)*(dx/dt) instead of dxdt/dt, if I recall correctly, you cannot treat dt or dx as a variable, so they don't...
Homework Statement
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Homework Statement
This is a chain rule problem that I can't seem to get right no matter what I do. It wants me to find the derivative of y=sqrt(x+sqrt(x+sqrt(x)))
Homework Equations
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The Attempt at a Solution
My...
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Homework Statement
Evaluate the derivative of the following function:
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Homework Equations
Chain Rule
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Hi,
I have a probably very stupid question:
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Homework Statement
(a) Light waves satisfy the wave equation ##u_{tt}c^2u_{xx}## where ##c## is the speed of light.
Consider change of coordinates $$x'=xVt$$ $$t'=t$$
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Homework Statement
here is the problem statement
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Homework Equations
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The Attempt at a Solution
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Homework Statement
a. Given u=F(x,y,z) and z=f(x,y) find { f }_{ xx } in terms of the partial derivatives of of F.
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Homework Equations
Implicit function theorem, chain rule diagrams, Clairaut's...
Suppose you have a parameterized mulivaried function of the from ##F[x(t),y(t),\dot{x}(t),\dot{y}(t)]## and asked to find ##\frac{dF}{dt}##, is this the correct expression according to chain rule? I am confused because of the derivative terms involved.
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Homework Statement
Use the Chain Rule to find the 1. order partial derivatives of g(s,t)=f(s,u(s,t),v(s,t)) where u(s,t) = st & v(s,t)=s+t
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Homework Statement
Homework Equations
Chain rule, partial derivation
The Attempt at a Solution
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Homework Statement
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Homework Equations
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