Asked by Cupcake
I do not understand implicit differentiation. One of the problems are: find y' by implicit differentiation xy+2x+3x^2=4.
I would appreciate any help that can be offered.
I would appreciate any help that can be offered.
Answers
Answered by
Reiny
if you can do the basic rules like the product rule, then you can do implicit differentiating.
xy+2x+3x^2=4
Let me explain term by term
xy is a product, so when you differentiate
you get
x(dy/dx) + y(dx/dx)
or
x(dy/dx) + y since dx/dx = 1
2x would give you 2 and
3x^2 would give you 6x (really it was 6x(dx/dx) which is 6x(1) or 6x)
and of course the derivative of 4 is zero
so you would have
x(dy/dx) + y + 2 + 6x = 0
solving this for dy/dx you would finally have
dy/dx = (-6x - y - 2)/x
You will get questions where there is a dy/dx in several terms.
In that case bring all those terms to one side of your equation, factor out the dy/dx, and solve for dy/dx that way.
the key this is to remember:
if you differentiate an x term you also get a dx/dx hanging around, which is 1, so you don't have to write it, but
if you differentiate a y term you would get a dy/dx hanging around.
xy+2x+3x^2=4
Let me explain term by term
xy is a product, so when you differentiate
you get
x(dy/dx) + y(dx/dx)
or
x(dy/dx) + y since dx/dx = 1
2x would give you 2 and
3x^2 would give you 6x (really it was 6x(dx/dx) which is 6x(1) or 6x)
and of course the derivative of 4 is zero
so you would have
x(dy/dx) + y + 2 + 6x = 0
solving this for dy/dx you would finally have
dy/dx = (-6x - y - 2)/x
You will get questions where there is a dy/dx in several terms.
In that case bring all those terms to one side of your equation, factor out the dy/dx, and solve for dy/dx that way.
the key this is to remember:
if you differentiate an x term you also get a dx/dx hanging around, which is 1, so you don't have to write it, but
if you differentiate a y term you would get a dy/dx hanging around.
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