Implicit Differentiation (AQA A-Level Mathematics): Revision Notes

7.5.1 Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function when it is not explicitly solved for one variable in terms of the other. Instead of solving the equation for y in terms of $x$ first, you differentiate both sides of the equation with respect to $x$, treating $y$ as a function of $x$.

1. When to Use Implicit Differentiation:

• Implicit Equations: Implicit differentiation is used when the equation is given in a form where $y$is not isolated on one side (e.g., $x^2 + y^2 = 1, x^3 + y^3 = 3xy$).
• Relations: It's often used in cases where $y$ is implicitly defined as a function of $x$ through some relation.
• Circular Functions: For curves like circles or ellipses (e.g.$, x^2 + y^2 = r^2).$

2. Steps for Implicit Differentiation:

3. Example Problems:

4. Why Implicit Differentiation Works:

Implicit differentiation works because it leverages the chain rule, treating $y$ as a function of $x$ (i.e., $y = g(x)$). When you differentiate a term involving $y$, you apply the chain rule, which introduces $\frac{dy}{dx}$ into the equation.

5. Applications of Implicit Differentiation:

• Tangent and Normal Lines: Finding the slope of the tangent or normal line to a curve at a given point.
• Related Rates: In related rates problems, implicit differentiation helps to relate the rates of change of multiple variables.
• Finding Derivatives of Inverse Functions: Implicit differentiation can also be used to find derivatives of inverse functions when the function itself is not easily invertible.

6. Higher-Order Derivatives:

Implicit differentiation can also be used to find higher-order derivatives (e.g., $\frac{d^2y}{dx^2}).$After finding $\frac{dy}{dx}$, differentiate again with respect to $x$, applying implicit differentiation as needed.

Given $x^2 + y^2 = 1$, we found $\frac{dy}{dx} = \frac{-x}{y}$. To find $\frac{d^2y}{dx^2}$:

• Differentiate $\frac{dy}{dx} = \frac{-x}{y}$ implicitly with respect to $x$:

$\frac{d^2y}{dx^2} = \frac{-y - (-x) \cdot \frac{dy}{dx}}{y^2} = \frac{-y + x \cdot \frac{x}{y}}{y^2} = \frac{-y + \frac{x^2}{y}}{y^2} = \frac{-y^2 + x^2}{y^3}$

Summary:

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Implicit Differentiation

An explicit function is one that can, for example, be written in the form $y = f(x)$, e.g., $y = x^2 + 7$.

An implicit function is not written in this way, e.g., $x^2 + y^2 + 5xy = \cos x$. This cannot be rearranged in the form $y = f(x)$.

Differentiating Implicitly

As long as we do the same thing to both sides of an equation, the fact that they are equal is maintained. This includes differentiating.

$$\frac{d}{dx} \text{ means "differentiate with respect to } x \text{"}$$ $$\frac{d}{dx} (x^2) = 2x, \quad \frac{d}{dx} (\tan x) = \sec^2 x, \quad \frac{d}{dx} (e^x) = e^x$$

However, a problem is encountered when trying to differentiate an expression W.R.T. $x$ that involves $y$.

Note:

$$\frac{d}{dx} \equiv \frac{d}{dy} \times \frac{dy}{dx}$$

This says that differentiation with respect to $x$ is exactly equivalent to differentiating with respect to $y$, then multiplying by $\frac{dy}{dx}$.

e.g.,

$$\frac{d}{dx} (y^2) \equiv \frac{d}{dy} (y^2) \times \frac{dy}{dx} = 2y \frac{dy}{dx}$$ $$\frac{d}{dx} (\cos y) \equiv \frac{d}{dy} (\cos y) \times \frac{dy}{dx} = -\sin y \frac{dy}{dx}$$

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Applying this to a function like $y = x^2 + 2x$ (an explicit function):

Differentiating both sides, we get:

$$\frac{d}{dx}(y) = \frac{d}{dx}(x^2 + 2x)$$ $$\frac{d}{dy}(y) \times \frac{dy}{dx} = 2x + 2$$ $$1 \times \frac{dy}{dx} = 2x + 2 \Rightarrow \frac{dy}{dx} = 2x + 2$$

(The result expected)

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Differentiating implicitly:

$$3x^2 + 8xy + 4x^2 \frac{dy}{dx}+ 3y^2 \frac{dy}{dx} = 0$$ $$4x^2 \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = -3x^2 - 8xy$$ $$\frac{dy}{dx}(4x^2 + 3y^2) = -3x^2 - 8xy$$ $$\frac{dy}{dx} = \frac{-3x^2 - 8xy}{4x^2 + 3y^2}$$

At the point $(1, 1)$:

$$\frac{dy}{dx} = \frac{-3(1)^2 - 8(1)(1)}{4(1)^2 + 3(1)^2} = \frac{-3 - 8}{4 + 3} = \frac{-11}{7}$$

Gradient of normal:

$$= -\frac{1}{\frac{-11}{7}} = \frac{7}{11}$$

Point: (1, 1)

$$y - 1 = \frac{7}{11}(x - 1)$$ $$11y - 11 = 7(x - 1)$$ $$11y - 11 = 7x-7$$ $$:success[7x - 11y + 4 = 0]$$

(or any integer multiple of this)

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Differentiating both sides W.R.T. $x$:

$$y^2 + 2xy\frac{dy}{dx} = 2x$$ $$2xy\frac{dy}{dx} = 2x - y^2$$ $$\frac{dy}{dx} = \frac{2x - y^2}{2xy}$$

Setting $\frac{dy}{dx} = 0$ to find stationary points:

$$\frac {2x - y^2}{2xy} = 0 \Rightarrow 2x-y^2 = 0\Rightarrow x = \frac{y^2}{2}$$ $$\text{Stationary points lie on the curve } x = \frac{y^2}{2} \text{ and the original curve.}$$

Substituting $x = \frac{y^2}{2}$ into the original curve equation:

$$y^2 \left(\frac{y^2}{2}\right) = \left(\frac{y^2}{2}\right)^2 + 1$$ $$\frac{y^4}{2} = \frac{y^4}{4} + 1$$ $$\frac{y^4}{4} = 1 \Rightarrow y^4 = 4 \Rightarrow y =\pm\sqrt{\sqrt{4}}=\pm\sqrt{2}$$

Substituting $y = \pm\sqrt{2}$ into $x = \frac{y^2}{2}$:

$$y = \sqrt{2} \Rightarrow x = \frac{(\sqrt{2})^2}{2} = 1$$ $$y = -\sqrt{2} \Rightarrow x = \frac{(-\sqrt{2})^2}{2} = 1$$ $$\text{Stationary points at } :success[(1, \sqrt{2})] \text{ and } :success[(1, -\sqrt{2})]$$

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