# Homework Help: Second derivative chain rule

1. Mar 17, 2017

### mastermechanic

1. The problem statement, all variables and given/known data
Question has been attached to topic.

2. Relevant equations
Chain rule.

3. 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!

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2. Mar 17, 2017

### 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.

3. Mar 18, 2017

### 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=tm, y=tn and it did not seem to work.

4. Mar 18, 2017

### John Park

Last edited: Mar 18, 2017
5. Mar 18, 2017

### haruspex

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.

6. Mar 18, 2017

### John Park

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