Derivative of a derivative with respect to something else

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In summary, the problem involves finding the second derivative of a function in terms of its first derivative and proving a relation involving the exponential function. The solution involves using the chain rule and algebraic manipulation to obtain the correct expression for the second derivative.
  • #1
jaderberg
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Homework Statement



If [tex]x=e^t[/tex], find [tex]\frac{dy}{dx}[/tex] in terms of [tex]\frac{dy}{dt}[/tex] and hence prove [tex]\frac{d^2y}{dx^2}=e\stackrel{-2t}{}(\frac{d^2y}{dt^2}-\frac{dy}{dt})[/tex]

Homework Equations

The Attempt at a Solution



Well i have done the first bit:

[tex]x=e^t[/tex]

[tex]\frac{dx}{dt}=x[/tex]

[tex]\frac{dy}{dx}=x\stackrel{-1}{}\frac{dy}{dt}[/tex]

Thats fine so now to get the second derivative using product rule:

[tex]\frac{d^2y}{dx^2}=-x\stackrel{-2}{}[/tex][tex]\frac{dy}{dt}+x\stackrel{-1}{}[/tex][tex]\frac{d(\frac{dy}{dt})}{dx}[/tex]

now my problem is what is [tex]\frac{d(\frac{dy}{dt})}{dx}[/tex]?

would greatly appreciate a step by step for that part so i can finally understand this one!
 
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  • #2
consider finding [tex]\frac{d}{dx}(y)[/tex]

[tex]\frac{d}{dx}=\frac{d}{dy} \frac{dy}{dx}[/tex]

so now you will have
[tex]\frac{d}{dx}(y)=\frac{d}{dy}(y) \frac{dy}{dx}=1\frac{dy}{dx}[/tex]
 
  • #3
jaderberg said:
...now my problem is what is [tex]\frac{d(\frac{dy}{dt})}{dx}[/tex]?

would greatly appreciate a step by step for that part so i can finally understand this one!

You again invoke chain rule:

[tex]\frac{dx}{dt}\cdot \frac{d(\frac{dy}{dt})}{dx}= \frac{d(\frac{dy}{dt})}{dt}=\frac{d^2 y}{dt^2}[/tex]

That and a little algebra should get you home.
 
  • #4
im sorry but i still don't understand how to do it...
rockfreak: what happens now when i change y in your expression to [tex]\frac{dy}{dt}[/tex]?

jam: if that was true then i would get [tex]\frac{d^2y}{dx^2}=-x\stackrel{-2}{}\frac{dy}{dt}+x\stackrel{-1}{}\frac{d^2y}{dt^2}[/tex]

which would produce the wrong answer...
 
  • #5
ok sorry i think i got it...correct me if I am wrong:

i could write [tex]\frac{d(\frac{dy}{dt})}{dx}=\frac{dt}{dx}\frac{d(\frac{dy}{dt})}{dt}=x\stackrel{-1}{}\frac{d^2y}{dt^2}[/tex]

thanks guys!
 
  • #6
jaderberg said:
rockfreak: what happens now when i change y in your expression to [tex]\frac{dy}{dt}[/tex]?

You still follow the same rules...in fact, you could use the convention from Newtonian mechanics that a dot over a quantity represents its time derivative:

[tex] \dot y = \frac{dy}{dt} [/tex]

To make things less cumbersome. However, rockfreak essentially wrote down that:

[tex] \frac{d \dot y}{dx} = \frac{d \dot y}{dx} [/tex]

which isn't that useful. Jambaugh had the correct advice

jaderberg said:
jam: if that was true then i would get [tex]\frac{d^2y}{dx^2}=-x\stackrel{-2}{}\frac{dy}{dt}+x\stackrel{-1}{}\frac{d^2y}{dt^2}[/tex]
which would produce the wrong answer...

Nope, you would not get that. Check it again. You confused x with t at some point.


Let's go over jambaugh's method for clarity. Using the chain rule, and keeping in mind that the dot denotes differentiation w.r.t. time, we can write:

[tex] \frac{d}{dx}\left(\frac{dy}{dt}\right) = \frac{d \dot y}{dx} [/tex]

[tex] = \frac{d \dot y} {dt} \frac{dt}{dx} = \frac{d^2y}{dt^2} \frac{dt}{dx} [/tex]
 
  • #7
yeh cheers i think i got that...pretty simple actually once you know what to look for
 

1. What is a derivative of a derivative with respect to something else?

A derivative of a derivative with respect to something else, also known as a second-order derivative, is the rate of change of a rate of change. It measures how the rate of change of a quantity is itself changing.

2. How is a derivative of a derivative calculated?

A derivative of a derivative is calculated by taking the derivative of the original function with respect to the variable, and then taking the derivative of that result with respect to the same variable. This can be denoted as f''(x) or (d^2y)/(dx)^2.

3. What is the significance of the derivative of a derivative?

The derivative of a derivative provides information about the curvature of a function. A positive second-order derivative indicates a concave up shape, while a negative second-order derivative indicates a concave down shape. It can also be used to find points of inflection on a graph.

4. What is the difference between a first-order and second-order derivative?

A first-order derivative, also known as a simple derivative, measures the instantaneous rate of change of a function. A second-order derivative measures the rate of change of the first-order derivative, or the rate of change of the rate of change.

5. In what fields is the concept of a derivative of a derivative commonly used?

The concept of a derivative of a derivative is commonly used in fields such as physics, engineering, economics, and statistics. It is also used in advanced calculus and in the study of differential equations.

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