# Chain rule Definition and 32 Discussions

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)}
and

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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.

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1. ### I Second derivative, chain rules and order of operations

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2. ### The below solution seems to assume that 1/0 = 0

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3. ### Derivatives and the chain rule

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5. ### Higher order derivatives using the chain rule

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6. ### I Divergence with Chain Rule

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7. ### I Chain rule problem

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8. ### A Why does MTW keep calling the "product rule" the "chain rule"?

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9. ### Help explaining the chain rule please

I had already calculated the first partial derivative to equal the following: $$\frac{\partial y}{\partial t} = \frac{\partial v}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial v}{\partial t}$$ Now the second partial derivative I can use the chain rule to do and get to...

15. ### Derivative of x(t)?

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...
16. ### Demystifying the Chain Rule in Calculus - Comments

Greg Bernhardt submitted a new PF Insights post Demystifying the Chain Rule in Calculus Continue reading the Original PF Insights Post.
17. ### Thermal Energy Equation Term - Chain Rule

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18. ### Chain rule in thermodynamics

I'm trying to understand why $$\left(\frac{\partial P}{\partial T}\right)_V = -\left(\frac{\partial P}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_P$$ where does the minus sign come from?
19. ### Chain rule problem

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 dy/dx=(dy/du)*(du/dx) d/dx sqrtx=1/(2sqrtx) d/dx x=1 (f(x)+g(x))'=f'(x)+g'(x) The Attempt at a Solution My...
20. ### I Rigorously understanding chain rule for sum of functions

In my quest to understand the Euler-Lagrange equation, I've realized I have to understand the chain rule first. So, here's the issue: We have g(\epsilon) = f(t) + \epsilon h(t). We have to compute \frac{\partial F(g(\epsilon))}{\partial \epsilon}. This is supposed to be equal to \frac{\partial...
21. ### Simplifying this derivative....

Homework Statement Evaluate the derivative of the following function: f(w)= cos(sin^(-1)2w) Homework Equations Chain Rule The Attempt at a Solution I did just as the chain rule says where F'(w)= -[2sin(sin^(-1)2w)]/[sqrt(1-4w^(2)) but the book gave the answer as F'(w)=(-4w)/sqrt(1-4w^(2))...

24. ### Derivative in spherical coordinates

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25. ### Partial derivatives and chain rule

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26. ### I Chain rule in a multi-variable function

Suppose you have a parameterized muli-varied 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. ##\frac{dF}{dt}=\frac{\partial...
27. ### Solve first order partial derivatives

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28. ### I Chain rule confusion

while solving differential equations, I got a bit confused with chain rule problem. The solution says below yprime = z then y double prime = z (dz/dy) = z prime but I don't understand why the differentiation of z is in that form. Please help...
29. ### Chain rule partial derivative

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30. ### Calc BC derivative problem with trig and double angle -- Help please

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31. ### Symmetry in second order partial derivatives and chain rule

When can I do the following where ##h_{i}## is a function of ##(x_{1},...,x_{n})##? \frac{\partial}{\partial x_{k}}\frac{\partial f(h_{1},...,h_{n})}{\partial h_{m}}\overset{?}{=}\frac{\partial}{\partial h_{m}}\frac{\partial f(h_{1},...,h_{n})}{\partial x_{m}}\overset{\underbrace{chain\...
32. ### Am I applying the chain rule correctly?

So i have an equation problem that i need to find the 2nd derivative of, but my understanding of the chain rule is not the best. I tried working it out but i don't know if i did it correctly. i was given the equation y=4(x2+5x)3 So to take the first derivative, i started off by using the chain...