DIFFERENTIATION 11
Implicit Differentiation
Implicit differentiation is used when \(y\) is defined by an equation involving both \(x\) and \(y\), rather than being isolated explicitly.
The main idea
Sometimes a relationship is written as
instead of
\[ y=f(x). \]We can still differentiate with respect to \(x\), remembering that \(y\) depends on \(x\).
1. Explicit versus implicit form
An explicit function gives \(y\) directly:
\[ y=x^2+1. \]An implicit relationship may look like
\[ x^2+y^2=25. \]Here \(y\) is not isolated.
2. Differentiate both sides
Consider
\[ x^2+y^2=25. \]Differentiate both sides with respect to \(x\):
\[ \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25). \]This gives
\[ 2x + \frac{d}{dx}(y^2) = 0. \]3. Why the chain rule appears
Because \(y\) depends on \(x\), \(y^2\) is really a composite function.
Using the chain rule,
\[ \frac{d}{dx}(y^2) = 2y\frac{dy}{dx}. \]So
\[ 2x + 2y\frac{dy}{dx} = 0. \]4. Solve for \(dy/dx\)
Rearrange:
\[ 2y\frac{dy}{dx} = -2x. \]Therefore,
This gives the slope of the circle at any point where \(y\neq0\).
5. Example at a particular point
For the circle
\[ x^2+y^2=25, \]consider the point
\[ (3,4). \]Then
\[ \frac{dy}{dx} = -\frac{3}{4}. \]So the tangent gradient at \((3,4)\) is \(-3/4\).
6. Another example
Consider
\[ x^2+xy+y^2=7. \]Differentiate with respect to \(x\):
\[ 2x + \frac{d}{dx}(xy) + 2y\frac{dy}{dx} = 0. \]Use the product rule on \(xy\):
\[ \frac{d}{dx}(xy) = y+x\frac{dy}{dx}. \]Therefore,
\[ 2x+y +x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0. \]7. Collect the derivative terms
Group the terms containing \(dy/dx\):
\[ \left(x+2y\right)\frac{dy}{dx} = -(2x+y). \]Therefore,
8. Why every derivative of \(y\) gains \(dy/dx\)
If \(y=y(x)\), then differentiating a function of \(y\) with respect to \(x\) requires the chain rule.
For example,
\[ \frac{d}{dx}(y^3) = 3y^2\frac{dy}{dx}, \] \[ \frac{d}{dx}(\sin y) = \cos y\frac{dy}{dx}, \] \[ \frac{d}{dx}(e^y) = e^y\frac{dy}{dx}. \]The factor \(dy/dx\) appears because \(y\) itself changes with \(x\).
9. Example with a trigonometric term
Suppose
\[ x+\sin y=2. \]Differentiate:
\[ 1+\cos y\frac{dy}{dx}=0. \]Therefore,
\[ \boxed{ \frac{dy}{dx} = -\frac{1}{\cos y} } \]10. Why implicit differentiation is useful
Some curves cannot easily be written as one single function \(y=f(x)\).
For example,
\[ x^2+y^2=25 \]describes a full circle.
Solving explicitly gives
\[ y=\pm\sqrt{25-x^2}, \]which separates the upper and lower halves.
Implicit differentiation lets us work directly with the original curve.