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fp2 help substiution in second order equations

fp2.jpgcan someone please help how do i get d2y/dx2 if ive got (cosx.dy/dx)=dy/dx
Reply 1
You will have to use implicit differentiation for questions like these. Not sure if what you got is right though? How did you do it?
Reply 2
Original post by B_9710
You will have to use implicit differentiation for questions like these. Not sure if what you got is right though? How did you do it?


Sorry i meant to say dy/dx=cosx.dy/dz how would i proceed to find d2y/dx2
Reply 3
Ok. That's looks right (I'm trusting you here :smile:). So we need to differentiate both sides with respect to x. To differentiate the RHS we need to use the product rule.
So we have d2ydx2=ddx(cosx)dydz+cosxddx(dydz) \displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx} \left (\cos x \right ) \frac{dy}{dz} + \cos x \frac{d}{dx} \left (\frac{dy}{dz} \right ) .
I assume your trouble is differentiating the second term.
ddx(dydz)=dzdxddz(dydz)=dzdxd2ydz2 \displaystyle \frac{d}{dx} \left (\frac{dy}{dz} \right ) = \frac{dz}{dx} \frac{d}{dz} \left (\frac{dy}{dz} \right ) = \frac{dz}{dx} \frac{d^2y}{dz^2} .
From the substitution you have been given you should be able to find an expression for dz/dx.
Hope this helps.
(edited 7 years ago)
Reply 4
Original post by B_9710
Ok. That's looks right (I'm trusting you here :smile:). So we need to differentiate both sides with respect to x. To differentiate the RHS we need to use the product rule.
So we have d2ydx2=ddx(cosx)dydz+cosxddx(dydz) \displaystyle \frac{d^2y}{dx^2}=\frac{d}{dx} \left (\cos x \right ) \frac{dy}{dz} + \cos x \frac{d}{dx} \left (\frac{dy}{dz} \right ) .
I assume your trouble is differentiating the second term.
ddx(dydz)=dzdxddz(dydz)=dzdxd2ydz2 \displaystyle \frac{d}{dx} \left (\frac{dy}{dz} \right ) = \frac{dz}{dx} \frac{d}{dz} \left (\frac{dy}{dz} \right ) = \frac{dz}{dx} \frac{d^2y}{dz^2} .
From the substitution you have been given you should be able to find an expression for dz/dx.
Hope this helps.


Thank you so much! Do you mind if over the next couple of days I send any more questions I get stuck on to you . Your explanation was very helpful
Reply 5
Original post by Cookie2314
Thank you so much! Do you mind if over the next couple of days I send any more questions I get stuck on to you . Your explanation was very helpful


That would be fine.
Reply 6
m1.jpg m1 im confused about part a-how is reaction force of the lift equal to the downward forces of albert and bella, surely you can only use the masses for the downwards force

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