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Example · Example 5

Q.If x2+y2=25x^2+y^2=25, find dydx\dfrac{dy}{dx}.

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Concept understanding — Implicit Differentiation

Implicit Differentiation

When yy isn't alone

You can differentiate y=x2+3xy=x^2+3x term by term because yy is written explicitly in terms of xx. But an equation like x2+y2=25x^2+y^2=25, or x3+y3=6xyx^3+y^3=6xy, does not give yy by itself — solving for yy is messy or downright impossible.

Implicit differentiation finds dydx\dfrac{dy}{dx} without isolating yy: treat yy as an unknown function of xx, differentiate the whole equation as it stands, then solve for dydx\dfrac{dy}{dx}.

The one key move: yy is really y(x)y(x)

Wherever yy appears, picture y(x)y(x) hiding inside. Differentiating a yy-term therefore needs the chain rule, which tacks on a factor of dydx\dfrac{dy}{dx}:

ddx(y2)=2y dydx.\frac{d}{dx}\big(y^2\big)=2y\,\frac{dy}{dx}.

That extra dydx\dfrac{dy}{dx} on every yy-term is the whole trick.

The procedure

  1. Differentiate both sides with respect to xx, treating yy as y(x)y(x).
  2. Each time you differentiate a yy-term, multiply by dydx\dfrac{dy}{dx} (chain rule); use the product rule on mixed terms such as xyxy.
  3. Gather all dydx\dfrac{dy}{dx} terms on one side, everything else on the other.
  4. Factor out dydx\dfrac{dy}{dx} and divide.

Worked example

For x2+y2=25x^2+y^2=25:

2x+2ydydx=0⇒dydx=−xy.2x+2y\frac{dy}{dx}=0 \quad\Rightarrow\quad \frac{dy}{dx}=-\frac{x}{y}.

The answer naturally contains both xx and yy — that is normal here. To get the slope at a point on the curve, substitute the coordinates after differentiating; there is no need to solve for yy first. …

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