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
f
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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.
I'm looking at one step in my thermodynamics book and they go from
pV = \nu*R*T
to
p*dV + V*dp = \nu * R * dT
I think there's an application of the chain rule in here but I don't see exactly how it's working. Could someone show me the steps in between?
Thanks!
Homework Statement
I have a function z, and I need to find the derivative dz/dt "using the chain rule without substitution"
Homework Equations
z = x^{2}y^{3} + e^{y}\cos x
x = \log(t^{2})
y = \sin(4t)
The Attempt at a Solution
\frac{\mathrm{d} z}{\mathrm{d} t} =...
Hi I've spent a long time trying to understand this chain rule proof but I just can't get it...
I have attached 2 pictures: the second one is an intuitive chain rule proof that turns out to be bogus and the first is the correct proof. So I am trying to understand first of all what does the...
Homework Statement
z = cos(x^2 + 3y^2)
x = ucosv
y=usinv
find dz/dv
Homework Equations
The Attempt at a Solution
I think I can do these fairly well, but I'm a little unsure of the "protocol" for which variables to put back in. Sometimes (in this case) I can't really put...
What I know from the chain rule is that if y and u are differentiable with respect to x then dy/dx = (dy/du)*(du/dx)
Now, why is this example doesn't work:
y = x^2
u = c
then we have dy/dx = (dy/du) * (du/dx) = (dy/du) * 0 = 0 doesn't equal 2x
I want an answer irrelated to the chain...
Im new on the forum, so I hope you guys will have some patience with me :-)
I have a question about the chain rule and partial differential equations that I can't solve, it's:
Write the appropriate version of the chain rule for the derivative:
∂z/∂u if z=g(x,y), where y=f(x) and...
Homework Statement
I'm trying to figure out how to take grad(f(x(t)) where x(t) is a vector. Since it's part of a physics problem, it's assumed x(t) is in 3-dimensional space.
The Attempt at a Solution
My guess is that grad(f(x(t)) = ((∂f/∂x)(∂x/∂x),(∂f/∂x)(∂x/∂y),(∂f/∂x)(∂x/∂z)) but...
This is another problem than I've been stuck on for a long time and I tried reading and watching videos but I only find first order partial differentiation with more than two variables or higher order partial differentiation with only two variables. (I'm not calling f a variable but I am calling...
Chain rule or Substitution rule?
Homework Statement
It appears that a standard result in ODE is the following: if f(x,t) is smooth enough, then the solution \psi(t) to the initial value problem:
x(t)=x_{0}+\int_{0}^{t}f(x(s),s)ds
x(0)=x_{0}
is continuously differentiable with respect to...
I'm trying to understand this one derivation but this one part keeps messing me up;
theta = tan^-1 (y/x)
r^2 = x^2 + y^2
d theta/ d x = y/ (x^2 + y^2) how did they get this line?
->ds/dt where s is the arc length in cartesian coordinates is ((dx/dt)^2+(dy/dt)^2)^(1/2).
-> Therefore by the chain rule ds/dt = ds/dp * dp/dt, but if I substitute dx/dt=dx/dp* dp/dt and dy/dt= dy/dp* dp/dt in the formula above, I get ds/dt=ds/dp * |dp/dt|??
What is happening?
->Even by...
I think i found the solution to my problem but i was hoping to have someone check to make sure i did not make a mistake.
\xi = x - ct...... (1)
u(t,x) = v(t,\xi)......(2)
Taking the derivative
d[u(t,x) = v(t,\xi)]
\frac{\partial u}{\partial t}dt + \frac{\partial u}{\partial x}dx =...
Homework Statement
I am trying to solve the transport PDE using a change of variables and the chain rule, and my problem seems to be with the chain rule. The PDE is:
\frac{\partial u}{\partial t}+c\frac{\partial u}{\partial x} = 0 ......(1)
The change of variables (change of reference frame)...
Hello,
Looking through a book on calculus I found the following explanation for the chain rule and I have one unclear thing that I'd like to ask for help on.
The canonical example is used, y is a function of u: y = u^{n} and u is a function of x (let's say) u = 3x - 2 therefore by composition...
Homework Statement
z = f (x^2 + y^2)
prove using multi var chain rule that
y * dz/dx - x * dz/dy = 0
Homework Equations
The Attempt at a Solution
honestly i just need to no how to start it then I am sure i could figure the rest out
so i would find dz/dx and dz/dy then...
Homework Statement
Let f be a differentiable function of one variable, and let
z = f(x + 2y). Show that
2∂z/∂x − ∂z/∂y = 0
Homework Equations
Multi-variable chain rule
The Attempt at a Solution
I have no idea where to start with this, any advice would be greatly appreciated...
Homework Statement
I'm trying to follow my textbook on an application of the chain rule.
Two objects are traveling in elliptical paths given by the following parametric equation.
x1 = 4 cos t
x2 = 2 sin 2t
y1 = 2 sin t
y2 = 3 cos 2t
At what rate is the distance between the two...
Homework Statement
Differentiate the functions using chain rule. 2(x3 −1)(3x2 +1)4
Homework Equations
Chain Rule = f ' (g(x))g' (x)
The Attempt at a Solution
I don't know how to do using chain rule, but product rule is easier
So using product rule,
= f ' (x) g(x) + f (x)g'...
Homework Statement
[12] 3.Thetemperature T(x, y) at a point of the xy-plane is given by
T(x,y)= ye^(x^2).
A bug travels from left to right along the curve y = x^2
at a speed of 0.01m/sec. The bug
monitors T(x, y) continuously. What is the rate of change of T as the bug passes through...
Suppose I have
F(x,y) and y=y(t) and x=x(t)
Therefore,
Ft = Fx*xt + Fy*yt. Right?
Can I write
Ftt = (Fxx*xt + Fyy*yt)*xt + Fx*xtt + (Fxx*xt + Fyy*yt)*yt + Fy*ytt
?
Basically I'm trying to figure out the second derivative by chain rule.
If a function is given by u = u(T,v) how to use the chain rule to write how u changes with respect to T & v.
Please specify the steps involved.
i understand chain rule as \frac{du}{dx} = \frac{du}{dy} \frac{dy}{dx}
Homework Statement
If Z= F(x-y), show that Zx + Zy = 0
Homework Equations
The Attempt at a Solution
Suppose I let Q = x-y. Then, by chain rule,
Fx(Q) * 1 + Fy(Q) * -1. By identity, this statement must hold for all values x,y. In particular, it must hold for x=y. By x=y...
Homework Statement So there is an exercise in which I should "verify" the chain rule for some functions.
In other words to do it by substitution, then doing by the formula and checking if the results are the same. (and checking with the book`s answer too)
For a few of them, they just don`t...
I was wondering how to prove the multivariable chain rule
\frac{\mbox{d}z}{\mbox{d}t}=\frac{\partial z}{\partial y}\frac{\mbox{d}y}{\mbox{d}t}+\frac{\partial z}{\partial x}\frac{\mbox{d}x}{\mbox{d}t}
where z=z(x(t),y(t))
I don't really need an extremely rigorous proof, but a slightly...
Homework Statement
w= f(x,y)
x = u + v Verify that Wxx - Wyy = Wuv
y = u - v
Homework Equations
The Attempt at a Solution
I know how to find Wu or Wv but I have no idea on how to proceed to find the 2nd order derivative (or 3rd,4rth etc.. obviously). I...
Hi there,
My equation to solve is (xy+(x^2))dx + (-1)dy=0
For method of exact solutions, the partials are not equal to each other so I cannot use
exact solutions (reverse chain rule)
I don't know how to solve this
Homework Statement
Find the derivative of the following:
Homework Equations
Y= x^3(5x-1)^4
The Attempt at a Solution
4(3x^2(5x-1)^3)(4(3x^2(3(5x-1)^2)(2(5x-1)(5)
Just some general questions as I'm confused with when to use chain rule when not to.
For instance, to find the derivative of e^sqrt(x), the right answer is to use chain rule to get e^sqrtx*the derivative of sqrt(x). BUT, isn't there a formula that: d/dx K^x = In(K)*K^x? K for constant and x...
I did a derivative problem, but my book says that my answer is wrong.
f(x)=x2(x-2)4
I didn't see much use in the chain rule so I used the product rule.
x2(4(x-2)3) + (x-2)4(2x)
=4x2(x-2)3 + 2x(x-2)4
The book says that instead of this, the answer is ...
x2(4(x-2)3(1)) + (x-2)4(2x) =...
Homework Statement
http://images.calcchat.com/solutionart/etf5e/03/d/se03d01063.png
Homework Equations
The Attempt at a Solution
I get to the third row, but can't simplify (Sin2θ)(Cos2θ). I'm looking at the trigonometric double angle formulas, and still can't figure out how the...
Homework Statement
cot^2(Cos\theta)Homework Equations
chain rule
f prime (x) = f prime(g(x) * g prime (x)
The Attempt at a Solution
I am not sure if I am just inputting the wrong numbers into webassign or I am just missing and important trig derivative and just completely off of the boat...
I always get muddled when I'm dealing with chain rule of any degree of complexity and also when dealing with powers of trig. functions - this problem contains both:
find \frac{\partial n}{\partial A} and \frac{\partial n}{\partial D} of the following function...
I'm curious if there's a chain rule for the commutator (I'll explain what I mean) just like there's a product rule ([AB,C]).
So, say you have an operator, which can be expressed in terms of another operator, and we know the commutation relationship between x and another operator, y. I'll call...
Solve the following:
d/dt cos(theta)
d/dt t sin(theta)
d/dt r cos (theta)
d/dt r^2 (theta)
d/dt e^ (-3x)
d/dt (x^2 + y^2)
I would assume all by the second one are 0 since your solving for terms dt and not theta, x, y, or r... I don't think its right at all. I know it goes something...
Hi,
Say x=position, v=velocity, a=acceleration, t=time.
Thanks!
EDIT: I just realized that 2/x is not a constant and thus I shouldn't have treated it as a constant (taking the derivative of it as 0). However, I don't understand how to take the derivative with respect to t of it.
I want to find the second order derivative for f(x,y),x(u,v),y(u,v), f depends on x and y, and x and y depends on u and v. I'm trying to find \frac{{\partial^2 f}}{{\partial v \partial u}}This is what I did:
\frac{{\partial f}}{{\partial u}}=\frac{{\partial f}}{{\partial x}}\frac{{\partial...
I'm confident in my math ability, but how is it that by using the chain
rule...
W_{x_1 \rightarrow x_2} = \int^{x_2}_{x_1} m \frac{dv}{dt} dx
can be turned into
W_{x_1 \rightarrow x_2} = \int^{x_2}_{x_1} m \frac{dv}{dx} \frac{dx}{dt} dx = \int^{v_2}_{v_1}mv dv
?
I understand the...
From james stewart calculus Early Transcendentals.Before he states the proof he intoduced a property of differentiable funcion
My problem is how we defined \epsilon to be 0 when \Delta x=0
where this is not in the Domain.
Homework Statement
Use the chain rule, the derivative formula Dxsinu=cosuDxu, together with the identities
cosx=sin(\pi/2 -x) and sinx=cos(\pi/2 -x)
to obtain the fomula for Dxcosx.
Homework Equations
Chain rule: dy/dx=dy/du\cdotdu/dx
The Attempt at a Solution
For my second...
Homework Statement
Use implicit differentiation to find dy/dx
2x^3+x^2y-xy^3 = 2
Homework Equations
Chain Rule et al.
The Attempt at a Solution
My questions is this. When deriving something like xy^3, apply the product rule to get
1y^3 + x\frac{d}{dx}y^3
I am confused on...
Homework Statement
If u=f(x,y) where x=escost and y=essint
show that d2u/dx2+d2u/dy2 = e-2s[d2u/ds2+d2u/dt2
The Attempt at a Solution
i have no idea!
question though, do the partial derivitives have to be solved and expanded then just show that one side equals the other or can...
Homework Statement
\frac{d}{dx}(x+(x+sin^2(x))^3)^4
Homework Equations
Calc up to Chain Rule.
The Attempt at a Solution
Using product and chain rule I got:
\frac{dy}{dx}=4(x+(x+sin^2(x))^3)^3(1+3(x+sin^2(x))^2)(1+\frac{d}{dx}sin^2(x))
Then I calculated the derivative of sin^2(x)...
Homework Statement
I have proven in two ways (correctly) that the derivative of ln|x| = 1/x (note absolute value does vanish)
Now I open my textbook and see a general rule that \frac{d}{dx} ln (u) = \frac{u'}{u}
And the not so general derivative of |x| is \frac{d}{dx} |x| =...
Hello! I'm currently taking Mth 251 and have been working on this chain rule for a bit. It's kind of straight forward, there's just a lot of chains inside of chains and so forth. I think I have a solution, but I'm not entirely sure its correct.
Homework Statement
Find the derivative of...