Finding the Second Derivative Using the Chain Rule

[mastermechanic]

Homework Statement

Question has been attached to topic.

Homework Equations

Chain rule.

The Attempt at a Solution

$$\frac {dy}{dt} . \frac{dt}{dx} = \sqrt{t^2+1}.cos(π.t)$$
$$\frac{d^2y}{dt^2}.(\frac{dt}{dx})^2 = 2 $$
$$\frac{d^2y}{dt^2}.(t^2+1).cos^2(π.t)= 2 $$ and for the t=3/4,
$$\frac{d^2y}{dt^2}.\frac{25}{16}.\frac{1}{2} = 2 $$
$$\frac{d^2y}{dt^2} = \frac{64}{25}$$
$$\frac{dy}{dt} = \frac{8}{5}$$

I count the dt\dx as the function itself because it is the previous status of the function, I mean the function in the problem statement is a result of dt/dx.

Is my solution correct? Is my approach correct? If not , where am I wrong and how to solve?

Thank you!

 

[John Park]

Check your arithmetic on the last step. Also note that cos(3π/4) is negative but there's a sign ambiguity for (dt/dx), if you're getting it from your second equation.

 

[haruspex]

Can you justify $\frac{d^2y}{dx^2}=\frac{d^2y}{dt^2}\left(\frac{dt}{dx}\right)^2$?
I tried it with x=t^m, y=tⁿ and it did not seem to work.

 

[John Park]

Can you justify $d ^2 y /d x ^2 = d ^2 y /d t ^2 ( d t /d x ) ^2$ ? I tried it with $x=t^m, y=t^n$ and it did not seem to work.

Reference https://www.physicsforums.com/threads/second-derivative-chain-rule.908041/#post-5719418

Hmm: $\frac d {dx} = \left( \frac {dt} {dx} \right ) \frac d {dt}$ applied twice to y looked credible. But you're right: it doesn't seem to work for your case.

 

[haruspex]

Hmm: $\frac d {dx} = \left( \frac {dt} {dx} \right ) \frac d {dt}$ applied twice to y looked credible. But you're right: it doesn't seem to work for your case.

For typing ease, I'll use dot for d/dt and ' for d/dx.
$\dot y=y'\dot x$
Differentiating and applying the product rule:
$\ddot y=\dot x \frac d{dt}y'+y'\ddot x$
$=\dot x \frac {dx}{dt}y"+y'\ddot x$
$=\dot x^2y"+y'\ddot x$.
That checks out with my example.

 

[John Park]

Looks good. I can insist to myself that I've learned from the experience.