cat_2002_12
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Don't understand one part of mark scheme
it says
dy/dx=-2x^-3 dx/dt

and then says
d2y/dx2=-2x^-3d2x/dt2 + 6x^-4 (dx/dt)^2

I DO NOT UNDERSTAND WHY IT IS 6x^-4 (dx/dt)^2 for the last bit. Surely using the product rule it would just be 6x^-4 dx/dt

Please someone explain
Sorry if its ridiculously easy but I'm stuck
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sam1990sam
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using the product rule to differentiate implicitly gives for the last bit: 6x^-4 (dx/dt)(dx/dt)= 6x^-4(dx/dt)^2
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Chaoslord
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(Original post by sam1990sam)
using the product rule to differentiate implicitly gives for the last bit: 6x^-4 (dx/dt)(dx/dt)= 6x^-4(dx/dt)^2
exactly

because your not differentiating with respect to x you have to put the dx/dt in there
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cat_2002_12
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Ah okay
Thanks
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Swayum
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(Original post by cat_2002_12)
Don't understand one part of mark scheme
it says
dy/dx=-2x^-3 dx/dt

and then says
d2y/dx2=-2x^-3d2x/dt2 + 6x^-4 (dx/dt)^2

I DO NOT UNDERSTAND WHY IT IS 6x^-4 (dx/dt)^2 for the last bit. Surely using the product rule it would just be 6x^-4 dx/dt

Please someone explain
Sorry if its ridiculously easy but I'm stuck
I don't know what the guys above are talking about; if you were differentiating w.r.t. x, you would not get what the mark scheme has. I looked up the mark scheme and found the reason; it's not dy/dx, it's dy/dt!

\frac{dy}{dt} = -2x^{-3}\frac{dx}{dt}

The first bit of the product rule (udv/dx) gives us

-2x^{-3}\frac{d(\frac{dx}{dt})}{dt} = -2x^{-3}\frac{d^2x}{dt^2}

The second bit of the product rule (vdu/dx) gives us

\frac{dx}{dt}\frac{d(-2x^{-3})}{dt}

Now look at that, we're trying to differentiate x w.r.t. t. It wouldn't just go to 6x^-4, because that would be differentiating w.r.t. x, as shown below:

z = -2x^-3
dz/dx = 6x^-4

(note: I've let z = -2x^-3 to simplify \frac{d(-2x^{-3})}{dt} to \frac{dz}{dt})

In order to differentiate w.r.t t, we have to use the chain rule:

z = -2x^-3

We need dz/dt. But we know that

dz/dt = dz/dx * dx/dt

Well then, as dz/dx = 6x^-4;

\frac{dz}{dt} = 6x^{-4}\frac{dx}{dt}

In other words, the last bit of the product rule (vdu/dx) is

\frac{dx}{dt}\frac{d(-2x^{-3})}{dt} = \frac{dx}{dt}\frac{dz}{dt} = 6x^{-4}(\frac{dx}{dt})^2

Put this together and it works. Do ask if you're still confused about something.
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sam1990sam
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(Original post by Swayum)
I looked up the mark scheme and found the reason; it's not dy/dx, it's dy/dt!
True, did the question from scrap and overlooked that.
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cat_2002_12
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Ohhhhhh
Thanks so much
Got really confused lol
Yeah makes sense now
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