Differentiate simple functions and relations defined implicitly or…

The idea

Some curves are easier to describe implicitly or parametrically. For implicit relations, differentiate each term with respect to x, remembering that y is a function of x, so d/dx(yⁿ) = n y^(n−1) dy/dx by the chain rule. For example, differentiating x² + y² = 25 gives 2x + 2y dy/dx = 0, so dy/dx = −x/y. Products such as xy require the product rule, giving x dy/dx + y. For parametric curves, x and y are given in terms of a parameter t; the first derivative is dy/dx = (dy/dt)/(dx/dt), provided dx/dt ≠ 0. For example, if x = t² and y = t³, then dy/dx = 3t²/(2t) = 3t/2. Only the first derivative is required here, so you do not need to find d²y/dx² for these implicit or parametric cases.

What earns the marks

• Differentiates implicit terms in y with respect to x, attaching dy/dx through the chain rule.
• Applies the product rule to mixed terms such as xy when differentiating implicitly.
• Rearranges the differentiated equation to make dy/dx the subject, factorising where necessary.
• Finds dx/dt and dy/dt for parametric equations and forms dy/dx as their quotient.
• Substitutes a given point or parameter value to obtain a numerical gradient where required.

Common mistakes and how to put them right

• Differentiating y terms as if y were a constant.

  Put it right by including dy/dx for every differentiated y term.

• Forgetting the product rule on terms such as xy.

  Put it right by writing x dy/dx + y.

• Inverting the parametric quotient.

  Put it right by dividing dy/dt by dx/dt, not the other way round.

Exam tips

• Collect all dy/dx terms on one side before factorising to isolate the derivative.
• For parametric questions, write dx/dt and dy/dt separately before forming the quotient.
• Substitute the given coordinates or parameter value only after obtaining the general expression for dy/dx.