7.10 Implicit differentiation

AQA Edexcel OCR A OCR B (MEI)
Explicit and implicit functions

All functions discussed so far have been explicit functions, defined in terms of 1 variable only, such as in the form of \(y=f(x)\).

Implicit functions are defined in terms of more than 1 variable, such as in the form of \(f(x,y) = 0 \).

Implicit differentiation

  • Terms involving \(x\) only are differentiated normally with respect to \(x\).

  • Terms involving \(y\) only are differentiated with respect to \(y\), multiplied by \(\dfrac{dy}{dx} \).

  • Terms involving products of \(x\) and \(y\) are differentiated using the rules above and the product rule.

The resulting equation is rearranged to make \(\dfrac{dy}{dx} \) the subject.

Equations of tangents and normals

Once the gradient function \(\dfrac{dy}{dx} \) is found, the equation of the tangent or normal at any specific point on the curve can be found in the usual way.

Turning points

Turning points can be found in the usual way, by solving \(\dfrac{dy}{dx} = 0 \).

Depending on the gradient function, the solution to the equation \(\dfrac{dy}{dx} = 0 \) might be in the form \(y=f(x)\) or \(x=f(y)\).

In that case, substituting \(x\) or \(y\) in the original equation and solving will provide the \(x\)- or \(y\)-coordinate of the turning point(s).
Important
Implicit differentiation

  • Terms involving \(x\) only are differentiated normally with respect to \(x\).

  • Terms involving \(y\) only are differentiated with respect to \(y\), multiplied by \(\dfrac{dy}{dx} \).

  • Terms involving products of \(x\) and \(y\) are differentiated using the rules above and the product rule.

The resulting equation is rearranged to make \(\dfrac{dy}{dx} \) the subject.
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