Implicit FunctionsWatch

#1
How to solve Question 14?
0
#2
See Attached.
0
4 weeks ago
#3
Implicit diff is an application of chain rule.

we understand in this situation (Q14) that we might rearrange eqn to give y = f(x), but we needn't bother when differentiating m but we must treat y as a function of x.
Our target is to carry out the operation d/dx on each term of the eqn, but we cannot carry out that operation per se on a y-term.
instead, we carry out d/dy on a y-term, then times by dy/dx so that effectively the 'dy' terms cancel.

so for example, d/dx(cosy)
becomes
(dy/dx).d/dy(cosy)

= (-siny)(dy/dx)

Can you finish off this question, now?
0
#4
(Original post by begbie68)
Implicit diff is an application of chain rule.

we understand in this situation (Q14) that we might rearrange eqn to give y = f(x), but we needn't bother when differentiating m but we must treat y as a function of x.
Our target is to carry out the operation d/dx on each term of the eqn, but we cannot carry out that operation per se on a y-term.
instead, we carry out d/dy on a y-term, then times by dy/dx so that effectively the 'dy' terms cancel.

so for example, d/dx(cosy)
becomes
(dy/dx).d/dy(cosy)

= (-siny)(dy/dx)

Can you finish off this question, now?
I've already gone through that but the problem arises when differentiating in terms of x. Surely one must then have to put y=arcsinx and y= arcosx on the LHS and use the values on the formula booklet to differentiate. Edit: Wait, I may be seeing what you are trying to say. You've got to differentiate twice on function of y but doesn't that mean you are left with a multiple of d2y/dx2 though the solution only contains x terms.
Last edited by JudaicImposter; 4 weeks ago
0
#5
^Edited.
0
4 weeks ago
#6
Not necessarily.

You might try sqrt(2)sin(y+pi/4) as the Rsin(theta + alpha) form of the original.

Then dx/dy

then reciprocal

then find sqrt(2)cos(y+pi/4) in terms of x

Doing this, I've found dy/dx in terms of x only, then second diff is straight forward.
0
#7
(Original post by begbie68)
Not necessarily.

You might try sqrt(2)sin(y+pi/4) as the Rsin(theta + alpha) form of the original.

Then dx/dy

then reciprocal

then find sqrt(2)cos(y+pi/4) in terms of x

Doing this, I've found dy/dx in terms of x only, then second diff is straight forward.
How to find sqrt(2)cos(y+pi/4) in terms of x ? I'm not very good at this and not getting near the answer. Would prefer to see the solution, understand it quickly and move on eventhough this is not the common protocol on this forum.
0
4 weeks ago
#8
(Original post by JudaicImposter)
How to find sqrt(2)cos(y+pi/4) in terms of x ? I'm not very good at this and not getting near the answer. Would prefer to see the solution, understand it quickly and move on eventhough this is not the common protocol on this forum.
You know sqrt(2)sin(y+pi/4) = x. And you know how to relate cos to sin (use cos^2 + sin^2 = 1). Should be enough.
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