Differentiating implicitly defined functions, finding tangents to implicit curves.
So far, all the functions you have differentiated look like y=f(x) — y is given explicitly as a function of x. But many important curves cannot be written this way.
Consider the circle x2+y2=25. You could rearrange to y=±25−x2, but that gives two functions (the top and bottom halves), and the algebra quickly becomes ugly for more complex curves.
Implicit differentiation lets you find dxdy without rearranging at all. You differentiate both sides of the equation with respect to x, applying the chain rule to any term involving y. It is elegant, powerful, and appears regularly at A2.
This technique combines everything you have learnt — the chain rule, the product rule, and the power rule — into one method. In the next lesson, we use implicit differentiation to find tangent and normal lines to implicit curves.
An explicit function has the form y=f(x). You can read off y directly.
An implicit equation relates x and y without separating them. Examples:
For these, you cannot easily (or sometimes at all) write y=….
When you differentiate x2 with respect to x, you get 2x. Easy.
But what happens when you differentiate y2 with respect to x? Here y is itself a function of x, so the chain rule kicks in:
dxd(y2)=2y⋅dxdy
Compare:
| Expression | Differentiate w.r.t. x | Reason |
|---|---|---|
| x2 | 2x | Power rule directly |
| y2 | 2y⋅dxdy | Chain rule — y is a function of x |
| x3 | 3x2 | Power rule directly |
| y3 | 3y2⋅dxdy | Chain rule |
| sinx | cosx | Standard result |
| siny | cosy⋅dxdy | Chain rule |
| ey | ey⋅dxdy | Chain rule |
The golden rule: Every time you differentiate something involving y, you must multiply by dxdy. This is because y depends on x, so the chain rule applies.
Answer at least 3 of 4 correctly to complete this section.
Find dxdy for the circle x2+y2=25.
Differentiate each term with respect to x:
dxd(x2)+dxd(y2)=dxd(25)
2x+2ydxdy=0
Now solve for dxdy:
2ydxdy=−2x
dxdy=−yx
This result makes geometric sense. At the point (3,4) on the circle, the gradient is −43. The radius from the origin to (3,4) has gradient 34, and −43×34=−1 — the tangent is indeed perpendicular to the radius, as it should be for a circle.
Find dxdy given that x3+y3=9.
Differentiate:
3x2+3y2dxdy=0
dxdy=−y2x2
Find dxdy given that x2+3xy+y2=7.
Differentiate term by term:
dxd(3xy)=3xdxdy+3y
Putting it all together:
2x+3xdxdy+3y+2ydxdy=0
Collect dxdy terms:
dxdy(3x+2y)=−2x−3y
dxdy=−3x+2y2x+3y
Watch out: When differentiating xy, you must use the product rule. The derivative of xy with respect to x is xdxdy+y, not just dxdy.
Find dxdy given that x2y+y3=8.
The term x2y is a product of x2 and y:
dxd(x2y)=x2dxdy+2xy(product rule)
dxd(y3)=3y2dxdy
So:
x2dxdy+2xy+3y2dxdy=0
dxdy(x2+3y2)=−2xy
dxdy=−x2+3y22xy
Answer at least 3 of 4 correctly to complete this section.
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