1
$\begingroup$

I circled the step that i have no idea how was done, please help me guys: enter image description here

What kind of chain rule is that? I know the chain rule when we have a fucntion of multiple variables and a curve, if the function is evaluated at the curve then I know how to calculate the chain rule ( By doing the dot product of the gradient of the function evaluated at the curve by the derivative of the curve ) In this case I do not see how that apllies at all, I can't even see how I would calculate the gradiant of this function since its coordinates are functions themselves ! I am lost guys, clarify this to me please.

$\endgroup$
1
$\begingroup$

Let $x_1=y+\alpha \eta$ and $x_2=y'+\alpha \eta'$. Then, we have

$$\begin{align} \frac{\partial f(y+\alpha \eta, y'+\alpha \eta',x)}{\partial \alpha}&=\frac{\partial f(x_1,x_2,x)}{\partial \alpha}\\\\ &=\frac{\partial f(x_1,x_2,x)}{\partial x_1}\frac{\partial x_1}{\partial \alpha}+\frac{\partial f(x_x,x_2,x)}{\partial x_2}\frac{\partial x_2}{\partial \alpha}+\frac{\partial f(x_1,x_2,x)}{\partial x}\frac{\partial x}{\partial \alpha}\\\\ &=\eta \,\frac{\partial f(x_1,x_2,x)}{\partial x_1}+\eta'\,\frac{\partial f(x_1,x_2,x)}{\partial x_2}\\\\ &=\eta \,\frac{\partial f(y+\alpha \eta,y'+\alpha \eta',x)}{\partial y}+\eta'\,\frac{\partial f(y+\alpha \eta,y'+\alpha \eta',x)}{\partial y'} \end{align}$$

$\endgroup$
  • $\begingroup$ What did you do at your last step? Why x1 turned into y and x2 turned into y' o.o ( I understood very well all the rest of it, and thank u for that. ) $\endgroup$ – Victor Jul 27 '17 at 19:12
  • $\begingroup$ We denoted $y+\alpha \eta$ by $x_1$ and denoted $y'+\alpha \eta'$ by $x_2$. So, taking the partial with respect to the first variable, call that $f_1$ is equivalent to $\frac{\partial f}{\partial x_1}=\frac{\partial f}{\partial y}$ since $x_1=y+\text{things independent of}\,y$. Similarly, taking the partial with respect to the second variable, call that $f_2$ is equivalent to $\frac{\partial f}{\partial x_2}=\frac{\partial f}{\partial y'}$ since $x_2=y'+\text{things independent of}\,y'$ $\endgroup$ – Mark Viola Jul 27 '17 at 19:16
  • $\begingroup$ I understand why we want to take the partial derivative with respect to the first variable, what I recognize as being the first variable is "y+αη" , why taking the derivative with respect to y+αη is the same as taking the derivative with respect to y ?? I do not understand why y is more special than alpha or eta in any way... ( Thank you anyway you are helping me so much. ) $\endgroup$ – Victor Jul 27 '17 at 19:24
  • $\begingroup$ Victor, $\alpha \eta$ does not depend on $y$. So, with respect to $y$, $\alpha \eta $ is a constant. So, taking a partial with respect to $y$ is the same as taking a partial with respect to $x_1=y+\alpha \eta$. $\endgroup$ – Mark Viola Jul 27 '17 at 19:25
  • $\begingroup$ You say : " taking a partial with respect to y is the same as taking a partial with respect to x1=y+αη" and I get it, my doubt is: Is taking a partial derivative with respect to x1 the same as taking a partial derivative with respect to y? The way you say it makes me think that y also does not depends on alpha, so that taking the derivative with respect to x1 is the same as taking the derivative with respect to alpha.eta, i'm so confused haha I do not see why y gets to be the special boy here intead of alpha or eta.. $\endgroup$ – Victor Jul 27 '17 at 19:32
1
$\begingroup$

It is indeed the chain rule. $\partial f / \partial y$ on the right hand side meaning "the derivative w.r.t. the first variable", ...

For a function with only a one-dimensional argument, the chain rule can be stated as $$\frac{\partial f(x(t))}{\partial t} = x'(t)\cdot \frac{\partial f}{\partial x}(x(t)),$$ where $x'(t)$ is the inner derivative, and $\frac{\partial f}{\partial x}(x(t))$ is the outer derivative evaluated at $x(t)$. The outer derviative is computed without the knowledge of $x(t)$.

Example: $f(z) = z^2, x(t) = \sin(t)$. Then $$\frac{\partial f(x(t))}{\partial t} = \underbrace{(\sin(t))'}_{x(t)'}\cdot \underbrace{(2z)}_{f'}\underbrace{(z = \sin(t))}_{(*)},$$ where $(*)$ is simply filling in the argument function $x(t)$ into the derivative of $f$, which was taken without knowledge of $x(t)$.

In 2 dimensions, we have $$\frac{\partial f(x(t), y(t))}{\partial t} = x'(t)\cdot \frac{\partial f}{\partial x}(x(t), y(t)) + y'(t)\cdot \frac{\partial f}{\partial y}(x(t), y(t)),$$ where $\frac{\partial f}{\partial x}(x(t), y(t))$ again means: We evaluate the derivative of $f$ with respect to its first variable/'slot', and then insert the specific arguments (here: $(x(t), y(t))$). This scheme is easily generalized to higher dimensions.

$\endgroup$
  • $\begingroup$ But the second variable is a function itself right? $\endgroup$ – Victor Jul 27 '17 at 19:02

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.