How do you find #(dy)/(dx)# given #y^2=4x#?

Answer 1

#dy/dx=2/y#

The equation is #y^2=4x#
When we differentiate the equation with respect to #x# we will apply the chain rule on the LHS.
#d/dxy^2=d/dx4x#
#2yd/dxy=4#
#2ydy/dx=4#
#dy/dx=2/y#
Here we know #y# is a variable, therefore we applied the chain rule.

If you want to check your answer ->

We found out that #dy/dx=2/y#

Now we will convert it into a differential equation

#ydy=2dx#

When we integrate (take the antiderivative) both the sides we get

#y^2/2=2x#
That is same as the equation we started out with i.e, #y^2=4x#
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Answer 2

To find (dy)/(dx) given (y^2 = 4x), differentiate both sides of the equation with respect to (x).

[\frac{d}{dx}(y^2) = \frac{d}{dx}(4x)]

Using the chain rule and power rule, we get:

[2y\frac{dy}{dx} = 4]

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

[\frac{dy}{dx} = \frac{4}{2y} = \frac{2}{y}]

Substitute (y^2 = 4x) to get:

[\frac{dy}{dx} = \frac{2}{\sqrt{4x}} = \frac{1}{\sqrt{x}}]

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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.

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.

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.

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.

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