Derivative Help
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Deleted member 4993
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Use chain rule:
dF/dt = dF/dx * dx/dt + dF/dy * dy/dt
Please share your work and tell us EXACTLY where you are stuck.
Steven G
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1st use equal signs instead of 'is'
What if you never were told that x(t)=-4t^4 and y(t)=6t and were just given F(x,y)=-6x^3-8y^5. How would you find dF/dx and DF/dy?? Please answer that question first.
Would I just take a second derivative of 4t^4 to get dx/dt?
So, dF/dx[4t^4]=16t^3
dx/dt[16t^3]=48t^2
?
BigBeachBanana
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I think you're so lost. Ignore t for now. Compute:
[math]\frac{\partial F}{\partial x}=\frac{\partial(6x^3-8y^5)}{\partial x}=?\\ \frac{\partial F}{\partial y}=\frac{\partial(6x^3-8y^5)}{\partial y}=?\\[/math]
If you are not asked to use the chain rule, you can do it directly by substituting the values of \(\displaystyle x\) and \(\displaystyle y\) in \(\displaystyle F\).
\(\displaystyle F(x,y)\) will become \(\displaystyle F(t)\).
\(\displaystyle F(t) = -6(-4t^4)^3 - 8(6t)^5 = 384t^{12} - 62208t^5\)
Now you can find \(\displaystyle \frac{dF}{dt}\) in one shot.
But if I were you, I would always use the chain rule. It is more fancy and it allows you to solve complicated functions without any headaches.
\(\displaystyle \frac{dF}{dt} = \frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt}\)
If this partial symbol \(\displaystyle \partial\) makes you confused, you can remove it and assume the derivative like this,
\(\displaystyle \frac{dF}{dt} = \frac{d F}{d x}\frac{dx}{dt} + \frac{d F}{d y}\frac{dy}{dt}\)
The brain is supposed to function better now
\(\displaystyle F(x,y)\) will become \(\displaystyle F(t)\).
\(\displaystyle F(t) = -6(-4t^4)^3 - 8(6t)^5 = 384t^{12} - 62208t^5\)
Now you can find \(\displaystyle \frac{dF}{dt}\) in one shot.
But if I were you, I would always use the chain rule. It is more fancy and it allows you to solve complicated functions without any headaches.
\(\displaystyle \frac{dF}{dt} = \frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt}\)
If this partial symbol \(\displaystyle \partial\) makes you confused, you can remove it and assume the derivative like this,
\(\displaystyle \frac{dF}{dt} = \frac{d F}{d x}\frac{dx}{dt} + \frac{d F}{d y}\frac{dy}{dt}\)
The brain is supposed to function better now
I am indeed so lost. So
So...If you are not asked to use the chain rule, you can do it directly by substituting the values of \(\displaystyle x\) and \(\displaystyle y\) in \(\displaystyle F\).
\(\displaystyle F(x,y)\) will become \(\displaystyle F(t)\).
\(\displaystyle F(t) = -6(-4t^4)^3 - 8(6t)^5 = 384t^{12} - 62208t^5\)
Now you can find \(\displaystyle \frac{dF}{dt}\) in one shot.
But if I were you, I would always use the chain rule. It is more fancy and it allows you to solve complicated functions without any headaches.
\(\displaystyle \frac{dF}{dt} = \frac{\partial F}{\partial x}\frac{dx}{dt} + \frac{\partial F}{\partial y}\frac{dy}{dt}\)
If this partial symbol \(\displaystyle \partial\) makes you confused, you can remove it and assume the derivative like this,
\(\displaystyle \frac{dF}{dt} = \frac{d F}{d x}\frac{dx}{dt} + \frac{d F}{d y}\frac{dy}{dt}\)
The brain is supposed to function better now
dF/dt= 4608t^11-311,040t^4??
This just seems wrong
Can you please show what it would like like doing the chain rule?
That's where I'm getting lost. I don't know how to do any of this...thank you for your help.
I don't know how to do this...would I take the derivative of each one so the first answer would be: 18x^2-8y^5?
And the second would be 6x^3-40y^5?
BigBeachBanana
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When you are computing partial derivative respect to x, treaty y as a constant and vice versa:
[math]\frac{\partial F}{\partial x}=\frac{\partial(6x^3-8y^5)}{\partial x}=18x^2\\ \frac{\partial F}{\partial y}=\frac{\partial(6x^3-8y^5)}{\partial y}=-40y^4\\[/math]Now, compute:
[math]\frac{dx}{dt}=\frac{d(-4t^4)}{dt}=?\\ \frac{dy}{dt}=\frac{d(6t)}{dt}=?\\[/math]Then, put it all together using chain rule:
[math]\frac{dF}{dt}=\frac{\partial F}{\partial x}\frac{dx}{dt}+ \frac{\partial F}{\partial y}\frac{dy}{dt}[/math]Lastly, convert [imath]x\,\&\, y[/imath] in terms of [imath]t[/imath] only.
BigBeachBanana
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You missed the negative. Understanding what derivatives mean will help you be less confused.
The interpretation of the derivative [imath]f'(x)[/imath] in a single variable case is the rate of change of the function as [imath]x[/imath] changes. The problem with multivariable functions is that there is more than one variable. In other words, what do we do if we only want one of the variables to change, or if we want more than one of them to change? If we’re going to allow more than one of the variables to change, there will be an infinite number of ways for them to change. That's why we treat other variables as constants when computing partial derivatives.
Now you have all the pieces of the puzzle, put them together.
BigBeachBanana
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Look at the original post you made, what does x equal and what does y equal?
If I am correct, you will have to buy me a bar of chocolate?
When you finish your final derivative, compare it with mine
D
Deleted member 4993
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F = 6*x^3 - 8*y^5
dF/dt = 3*6*x^2 * (dx/dt) - 8*5*y^4 * (dy/dt)
So:
dF/dt = 18 * (-4*t^4)^2 * (-4*4*t^3) - 40 *(6*t)^4 * (6)
= 18 * 16 * (-16) * t^8 *t^3 - 40*1296*6 * t^4 = ?
There could be "sign" mistake/s in my calculations - too early in the morning (yea - that's the ticket.....)But it's x(t)...and y(t) so I would want to plug in those equations for t?
BigBeachBanana
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Correct. The goal is to put everything in terms of t, hence the meaning of the notation [imath]\frac{dF}{\red{dt}}[/imath], i.e. how is F changed as t changes?