MHB Calculating Second Order Derivative of w at t=1

[Yankel]

Hello all,

I need to find the second order derivative of w by t, and to calculate it's value at t=1.

This is what I know about w, x and y.

\[w=ln(x+y)\]

\[x=e^{t}\]

\[y=e^{-t}\]The answer in the book is:

\[\frac{4}{(e^{t}+e^{-t})^{2}}\]

I got another answer and I don't know what I did wrong, my solution is attached as an image.

View attachment 1997

Would appreciate your help with it. Thank you.

P.S According to Maple I am correct

 

[lfdahl]

Your calculations are all the way through OK. It is just the very end of your equations, that needs a lift:

Try to calculate the difference:

\[(e^t+e^{-t})^2-(e^t-e^{-t})^2\]

what do you get?

 

[ThePerfectHacker]

Start with a function $f(x,y)$. Substitute two functions $x=g(t)$ and $y=h(t)$ to get the composite function $F(t) = f(g(t),h(t))$. The chain rule says that,
$$ \frac{dF}{dt} = \frac{\partial f}{\partial x}\frac{dx}{dt} + \frac{\partial f}{\partial y}\frac{dy}{dt} $$
Now when you compute the second derivative you use the product rule, so,
$$ \frac{d^2 F}{dt^2} = \frac{d}{dt} \left( \frac{\partial f}{\partial x} \right) \frac{dx}{dt} + \frac{\partial f}{\partial x}\frac{d^2 x}{dt^2} + \frac{d}{dt} \left( \frac{\partial f}{\partial y} \right) \frac{dy}{dt} + \frac{\partial f}{\partial y}\frac{d^2y}{dt^2} $$
To make things easier to follow let $A = \frac{\partial f}{\partial x}$, so,
$$ \frac{d}{dt}\left( \frac{\partial f}{\partial x}\right) = \frac{dA}{dt} = \frac{\partial A}{\partial x} \frac{dx}{dt} + \frac{\partial A}{\partial y} \frac{dy}{dt} = f_{xx} \frac{dx}{dt} + f_{xy}\frac{dy}{dt} $$
In a similar way if we have,
$$ \frac{d}{dt}\left( \frac{\partial f}{\partial y} \right) = f_{yx}\frac{dx}{dt} + f_{yy} \frac{dy}{dt} $$
Therefore our 2nd derivative formula is,
$$ \frac{d^2F}{dt} = (f_{xx} x' + f_{xy}y')x' + f_x \cdot x'' + (f_{yx}x' + f_{yy}y')y' + f_y \cdot y'' $$
This simplifies to, (recall that $f_{xy} = f_{yx}$),
$$ \frac{d^2 F}{dt} = f_{xx} (x')^2 + 2f_{xy}x'y' + f_{yy}(y')^2 + f_x\cdot x'' + f_y\cdot y'' $$
Now if you substitute your functions you will get your answer.

 

[Boromir]

Hello all,

I need to find the second order derivative of w by t, and to calculate it's value at t=1.

This is what I know about w, x and y.

\[w=ln(x+y)\]

\[x=e^{t}\]

\[y=e^{-t}\]The answer in the book is:

\[\frac{4}{(e^{t}+e^{-t})^{2}}\]

I got another answer and I don't know what I did wrong, my solution is attached as an image.

View attachment 1997

Would appreciate your help with it. Thank you.

P.S According to Maple I am correct

The derivative of $x+y$ is $e^t-e^{-t}$. Using the chain rule therefore, the derivative of w is $e^t-e^{-t}$ multiplied by $\frac{1}{x+y}$ which is equal to $\frac{e^t-e^{-t}}{e^t+e^{-t}}$ Then use quotient rule. x and y are just labels- easier to bypass them

 

[Yankel]

lfdahl,

what I get is: 2e^(-2t) where does it lead me ?

ThePerfectHacker, honestly, you lost me

and Boromir, isn't what I did identical to what you suggest ?

guys, I am quite confused, did I get it wrong or right ? I can't see what I did wrong here.

 

[lfdahl]

lfdahl,

what I get is: 2e^(-2t) where does it lead me ?

ThePerfectHacker, honestly, you lost me

and Boromir, isn't what I did identical to what you suggest ?

guys, I am quite confused, did I get it wrong or right ? I can't see what I did wrong here.

\[(e^t+e^{-t})^2-(e^t-e^{-t})^2 = (x+y)^2-(x-y)^2 = x^2+y^2+2xy-(x^2+y^2-2xy)\]

Can you continue?

 

[Yankel]

Right, I get 4, I see what you mean. But I can't see where my last move was incorrect. Can it be that both expressions are equal ?

Look at it more simply, if my first derivative is tanh(t), doesn't it mean that it's derivative should be 1-tanh(t) ? Even without calculating all the way ?

 

[lfdahl]

I would expect:

tanh'(t) = 1 - tanh²(t)

- and this is also the result you get.

 

[Yankel]

Thanks, you were very helpful, and I found my mistake thanks to you.

It is 1-tanh(t)^2 and I wrote 1-tanh(t) without the power, that's why Maple told me the expressions are not identical...