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How do you use implicit differentiation to find #(dy)/(dx)# given #1=3x+2x^2y^2#?

Answer 1

#dy/dx = (-3-4xy^2)/(4x^2y)#

Differentiate the equation through term by term. Remember you are differentiating with respect to #x# so when differentiating #x# we get:

#x -> 1#

but when differentiating with respect to #y# we get:

#y -> (dy)/(dx)#.

So:

#1=3x+2x^2y^2#

Differentiating gives:

#0=3+4xy^2 + 4x^2y (dy)/(dx)#

which we obtain using the chain and product rule on the #2x^2y^2# term. So we now re-arrange to get #(dy)/(dx)#:

#-> dy/dx = (-3-4xy^2)/(4x^2y)#

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Answer 2

Axel Alvarez

To find (\frac{{dy}}{{dx}}) using implicit differentiation for the equation (1 = 3x + 2x^2y^2), follow these steps:

Differentiate both sides of the equation with respect to (x).
Apply the chain rule where necessary.
Solve the resulting equation for (\frac{{dy}}{{dx}}).

The derivative of the equation (1 = 3x + 2x^2y^2) with respect to (x) yields:

[0 = 3 + 4xy^2\frac{{dy}}{{dx}}]

Solve for (\frac{{dy}}{{dx}}):

[\frac{{dy}}{{dx}} = -\frac{3}{{4xy^2}}]

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.