y is a hidden function of x, so the chain rule fires: Picture the circle x² + y² = 25. There's no single 'y = …' — the curve has a top and a bottom — yet it still has a gradient at every point.
The trick: pretend y is some function of x that we just haven't written out. Then differentiating a y-term needs the chain rule, which tacks on a dy/dx:
• d/dx(x²) = 2x (an x-term — normal).
• d/dx(y²) = 2y·(dy/dx) (a y-term — chain rule adds dy/dx).
So you differentiate every term with respect to x, and any term that contains y leaves behind a dy/dx factor.
IB-style question — differentiate each term
A curve is defined by x² + y² = 25.
Find dy/dx in terms of x and y.
Step by step
- Differentiate every term with respect to x. The y² term needs the chain rule, so it leaves a dy/dx.
- The right side 25 is constant, so its derivative is 0. Now isolate the dy/dx term.
- Divide by 2y to make dy/dx the subject.
Final answer
dy/dx = −x/y (the gradient at any point (x, y) on the circle).
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A term like xy needs the product rule first, then the chain rule: When x and y are multiplied (e.g. xy), differentiate it as a product: keep one, differentiate the other.
d/dx(xy) = (1)·y + x·(dy/dx) = y + x·dy/dx — the x-part differentiates to 1, the y-part leaves a dy/dx.
Once you have dy/dx in terms of x and y, find a gradient by substituting the point's coordinates, then build the tangent with y − y₁ = m(x − x₁).
IB-style question — product term, then gradient
A curve has equation x² + xy + y² = 7.
Find dy/dx, then the gradient at the point (1, 2).
Step by step
- Differentiate term by term. The xy needs the product rule; the y² needs the chain rule.
- Group the dy/dx terms on one side, everything else on the other.
- Make dy/dx the subject.
- Substitute x = 1, y = 2 to get the gradient at that point.
Final answer
dy/dx = −(2x + y)/(x + 2y); at (1, 2) the gradient is −4/5.
IB-style question — equation of the tangent
Using the gradient −4/5 at (1, 2) above, find the equation of the tangent line.
Step by step
- Use the point–gradient form with m = −4/5 and (x₁, y₁) = (1, 2).
- Expand and tidy into y = mx + c.
Final answer
y = −(4/5)x + 14/5 (equivalently 4x + 5y = 14).