Find Expression for $\frac{d^2y}{dx^2}$ in Terms of Derivatives

[Dethrone]

I'm going through lots and lots of past exams, so here are some more questions :D :

If $y=f(w)$ and $w=g(x)$ where $f$ and $g$ are twice differentiable functions find an expression for$ \frac{d^2y}{dx^2}$ in terms of $\frac{dy}{dw}$, $\frac{d^2y}{dw^2}$, $\frac{dw}{dx}$, and $\frac{d^2w}{dx^2}$.

I didn't get far in this one. I just rewrote the given expression:

$$\frac{d^2y}{dx^2}=\frac{d\d{y}{x}}{dx}=\frac{d\d{f(w)}{x}}{dx}$$

By the wording of the question, I have to express it in terms of ALL four derivatives. Any hints? (Smoking)

 

[I like Serena]

I'm going through lots and lots of past exams, so here are some more questions :D :

If $y=f(w)$ and $w=g(x)$ where $f$ and $g$ are twice differentiable functions find an expression for$ \frac{d^2y}{dx^2}$ in terms of $\frac{dy}{dw}$, $\frac{d^2y}{dw^2}$, $\frac{dw}{dx}$, and $\frac{d^2w}{dx^2}$.

I didn't get far in this one. I just rewrote the given expression:

$$\frac{d^2y}{dx^2}=\frac{d\d{y}{x}}{dx}=\frac{d\d{f(w)}{x}}{dx}$$

By the wording of the question, I have to express it in terms of ALL four derivatives. Any hints? (Smoking)

Chain rule says:
$$\d y x = \d y w \d w x$$
(Thinking)

 

[Dethrone]

Okay, so:

$$\frac{d^2y}{dx^2}=\frac{d\d{y}{x}}{dx}=\frac{d\left(\d{y}{w}\d{w}{t}\right)}{dx}$$

Not sure how I can include $\frac{d^2y}{dw^2}$ and $\frac{d^2w}{dx^2}$ too.

 

[chisigma]

I'm going through lots and lots of past exams, so here are some more questions :D :

If $y=f(w)$ and $w=g(x)$ where $f$ and $g$ are twice differentiable functions find an expression for$ \frac{d^2y}{dx^2}$ in terms of $\frac{dy}{dw}$, $\frac{d^2y}{dw^2}$, $\frac{dw}{dx}$, and $\frac{d^2w}{dx^2}$.

I didn't get far in this one. I just rewrote the given expression:

$$\frac{d^2y}{dx^2}=\frac{d\d{y}{x}}{dx}=\frac{d\d{f(w)}{x}}{dx}$$

By the wording of the question, I have to express it in terms of ALL four derivatives. Any hints? (Smoking)

If $$y=f(w)$$ and $$w=g(x)$$ where $$f~\&~g$$ are both twice differentiable then...

$$\frac{{d^2 y}}{{dx^2 }} = \frac{{d^2 y}}{{dw^2 }}\left( {\frac{{dw}}{{dx}}} \right)^2 + \frac{{dy}}{{dw}}\frac{{d^2 w}}{{dx^2 }}$$...

Kind regards

$\chi$ $\sigma$

 

[chisigma]

If $$y=f(w)$$ and $$w=g(x)$$ where $$f~\&~g$$ are both twice differentiable then...

$$\frac{{d^2 y}}{{dx^2 }} = \frac{{d^2 y}}{{dw^2 }}\left( {\frac{{dw}}{{dx}}} \right)^2 + \frac{{dy}}{{dw}}\frac{{d^2 w}}{{dx^2 }}$$...

The procedure to arrive at the result is an easy application of the chain rule. Starting from...

$\displaystyle \frac{d y}{dx} = \frac{dy}{d w}\ \frac{d w}{d x}\ (1)$

... and applying the chain rule You obtain...

$\displaystyle \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} (\frac{d y}{d w}\ \frac{d w}{d x}) = \frac{d^{2} y}{d w^{2}}\ (\frac{d w}{d x})^{2} + \frac{d y}{d w}\ \frac{d^{2} w}{d x^{2}}\ (2)$

The third derivative $\displaystyle \frac{d^{3} y}{d x^{3}}$ can be obtained from (2) in the same way as well as the derivatives of any order... the computation of the derivative of order 1001 is not difficult, only a bit tedious (Dull)...

Kind regards

$\chi$ $\sigma$

 

[Dethrone]

Thanks! Makes complete sense. :D

So the third derivative would be (I'm too lazy to take the last derivative...(Giggle) ):

$$\displaystyle \frac{d^{3} y}{d x^{3}}=\frac{d}{dx}\left[ \frac{d^{2} y}{d w^{2}}\ (\frac{d w}{d x})^{2} + \frac{d y}{d w}\ \frac{d^{2} w}{d x^{2}}\ \right]$$

I've think I read this before, but I forget. Anyone have any ideas as to why $\frac{d^2y}{dx^2}$ is not written $\frac{d^2y}{d^2x}$. You've also seemed to imply that $(dx)^2=(dx^2)$.

 

[Dethrone]

$\displaystyle \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} (\frac{d y}{d w}\ \frac{d w}{d x}) = \frac{d^{2} y}{d w^{2}}\ (\frac{d w}{d x})^{2} + \frac{d y}{d w}\ \frac{d^{2} w}{d x^{2}}\ (2)$

(Wait) Did you not mean?
$\displaystyle= \frac{d^{2} y}{d w^{2}}\ (\frac{d w}{d x}) + \frac{d y}{d w}\ \frac{d^{2} w}{d x^{2}}\ (2)$

I don't see the reason for the $\left(\frac{dw}{dx}\right)^2$ since by the product rule, our second function is $\frac{dw}{dx}$

 

[chisigma]

(Wait) Did you not mean?
$\displaystyle= \frac{d^{2} y}{d w^{2}}\ (\frac{d w}{d x}) + \frac{d y}{d w}\ \frac{d^{2} w}{d x^{2}}\ (2)$

I don't see the reason for the $\left(\frac{dw}{dx}\right)^2$ since by the product rule, our second function is $\frac{dw}{dx}$

Applying the chain rule You have...

$\displaystyle \frac{d}{dx} (\frac{dy}{dw}\ \frac{dw}{dx}) = \frac{d}{dx} (\frac{d y}{dw})\ \frac {dw}{dx} + \frac{dy}{d w}\ \frac{d}{d x} (\frac{dw}{dx})\ (1)$

Now look at the term...

$\displaystyle \frac{d}{dx} (\frac{d y}{dw})\ \frac {dw}{dx} = \frac{d^{2} y}{d w^{2}}\ \frac{d w}{d x}\ \frac{dw}{d x}\ (2)$

... and all Your doubts will be resolved...

Kind regards

$\chi$ $\sigma$

 

[Dethrone]

I see it now :D
Thanks, $\chi \sigma$, for the clear explanation (Yes)

 

[I like Serena]

I've think I read this before, but I forget. Anyone have any ideas as to why $\frac{d^2y}{dx^2}$ is not written $\frac{d^2y}{d^2x}$. You've also seemed to imply that $(dx)^2=(dx^2)$.

$\def\d{\mathrm{d}}$

In words, the second derivative is the change in the change of y, divided by the change in x, and again divided by the change in x.

We might write it as:
$$\begin{aligned}
f''(a) &= \lim_{h \to 0} \frac{f'(a+h)-f'(a)}{h} \\
&= \lim_{h \to 0} \frac{ \lim\limits_{k\to 0} \frac{f(a+h+k)-f(a+h)}{k} -\lim\limits_{k\to 0} \frac{f(a+k)-f(a)}{k} }{h} \\
&= \lim_{h \to 0} \lim_{k\to 0} \frac{ (f(a+h+k)-f(a+h)) - (f(a+k)-f(a)) }{h \cdot k} \\
&= \lim_{\d x \to 0} \lim_{\d x'\to 0} \frac{ \d(f(a+\d x)) - \d(f(a)) }{\d x \cdot \d x'} \\
&= \lim_{\d x \to 0} \lim_{\d x'\to 0} \frac{ \d(\d(f(a))) }{\d x \cdot \d x'}
\end{aligned}$$Or:
$$\frac{\d^2y}{\d x^2} = \frac{\d(\d(y))}{\d x \cdot \d x}$$
Do you see the power? (Wondering) (Muscle)
The power of Leibniz' notation that is. (Emo)

 

[Dethrone]

Ah, I see. So $dx^2=(dx)^2$ or $dx \cdot dx$. Some reason, I see it as $dx^2=d \cdot x^2$, which is why I was initially curious. (Smoking)

 

[I like Serena]

Ah, I see. So $dx^2=(dx)^2$ or $dx \cdot dx$. Some reason, I see it as $dx^2=d \cdot x^2$, which is why I was initially curious. (Smoking)

Yep!

Oh, and if you don't see the power of Leibniz yet, compare:
$$f(g(x))' = f'(g(x)) \cdot g'(x)$$
with:
$$\frac{df}{dx} = \frac{df}{dw} \cdot \frac{dw}{dx}$$

Or:
$$(f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))}$$
with:
$$\frac{df}{dx} = \frac {1}{\frac{dx}{df}}$$

(Rock)

 

[Dethrone]

Yes, I love Leibniz! :D One goes into calculus to write $\d{y}{x}$, not $f'(x)$...(Cool)

Now that you've written that the slope of a function is equal to the multiplicative inverse of its inverse, I will quietly leeway to another question that I was too lazy to start a new thread to ask.

If $f(x)=x^5$, how do I evaluate $\left(f^{-1}\right)^{'}(32)$ in two different ways? The standard way of doing it is to find $f^{-1}$, then differentiate it, then plug in 32. What is the other way to do it? I have it feeling it requires this $(f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))}$

 

[I like Serena]

Yes, I love Leibniz! :D One goes into calculus to write $\d{y}{x}$, not $f'(x)$...(Cool)

Now that you've written that the slope of a function is equal to the multiplicative inverse of its inverse, I will quietly leeway to another question that I was too lazy to start a new thread to ask.

If $f(x)=x^5$, how do I evaluate $\left(f^{-1}\right)^{'}(32)$ in two different ways? The standard way of doing it is to find $f^{-1}$, then differentiate it, then plug in 32. What is the other way to do it? I have it feeling it requires this $(f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))}$

Seems like a good feeling to me. (Nod)
Did you try it? (Wondering)

 

[Dethrone]

Evaluating $\left(f^{-1}\right)^{'}(32)$ the normal way gives me $\frac{1}{80}$. Then another way,
$$(f^{-1}(x))' = \frac{1}{f'(f^{-1}(x))}=\frac{1}{f'(\sqrt[5]{x})}=\frac{1}{5(x^{4/5})}$$
Hence,
$$(f^{-1}(32))'=\frac{1}{5(32^{4/5})}=\frac{1}{80}$$
Now that I think about it, would you consider this "another method" of evaluating $\left(f^{-1}\right)^{'}(32)?$

 

[I like Serena]

Looks good!

Now that I think about it, would you consider this "another method" of evaluating $\left(f^{-1}\right)^{'}(32)?$

I would. Wouldn't you? (Wondering)

 

[Dethrone]

Yes! (Dance)

 

[chisigma]

The procedure to arrive at the result is an easy application of the chain rule. Starting from...

$\displaystyle \frac{d y}{dx} = \frac{dy}{d w}\ \frac{d w}{d x}\ (1)$

... and applying the chain rule You obtain...

$\displaystyle \frac{d^{2} y}{d x^{2}} = \frac{d}{dx} (\frac{d y}{d w}\ \frac{d w}{d x}) = \frac{d^{2} y}{d w^{2}}\ (\frac{d w}{d x})^{2} + \frac{d y}{d w}\ \frac{d^{2} w}{d x^{2}}\ (2)$

The third derivative $\displaystyle \frac{d^{3} y}{d x^{3}}$ can be obtained from (2) in the same way as well as the derivatives of any order... the computation of the derivative of order 1001 is not difficult, only a bit tedious (Dull)...

The explicit expression of the third derivative is...

$\displaystyle \frac{d^{3} y}{d x^{3}} = \frac{d^{3} y}{d w^{3}}\ (\frac{d w}{d x})^{3} + 3\ \frac{d^{2} y}{d w^{2}}\ \frac{d w}{d x}\ \frac{d^{2} w}{d x^{2}} + \frac{d y}{d w}\ \frac{d^{3} w}{d x^{3}}\ (1)$

Kind regards

$\chi$ $\sigma$

 

[Dethrone]

Hi $\chi \sigma$,

I got something slightly different, let me know where I made the mistake:

$$\frac{d^3y}{dw^3}\left(\frac{dw}{dx}\right)^3+2\frac{dw}{dx}\frac{d^2y}{dw^2}+\frac{d^2y}{dw^2}\frac{dw}{dx}\frac{d^2w}{dx^2}+\frac{d^3y}{dx^3}\frac{dy}{dw}$$

EDIT: I missed a chain rule somewhere, you are right :D, because
$$\frac{d}{dx}\left[\left(\frac{dw}{dx}\right)^2\right]=2\frac{dw}{dx} \cdot \frac{d^2w}{dx^2}$$