fp1 jan 2007 q7

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

Reply 1

sam1990sam

5

using the product rule to differentiate implicitly gives for the last bit: 6x^-4 (dx/dt)(dx/dt)= 6x^-4(dx/dt)^2

Reply 2

Chaoslord

16

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

OP

Ah okay
Thanks

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;