Implicity Differentiation

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Prerequiste Material:

Explicit Differentiation

Chain Rule for Differentation (care of math.libretexts.org)

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Describing Implicit Differentiation:

The page is about the differentation of mathematical mappings which are not expressed explicitly. Explicit mappings are functions of the form $y=f\left(x\right)$. For example: $y=mx+c$, $y=x^3-cosx$, etc. In previous study this might have been called $y$ is the subject of the formula.

If the mapping is in terms of two vaiable and neither is expression explicitly in terms of the other then this is an implicit mapping. For example: $x^2+y^2=25$.

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Worked Example 1:

Differentiate the following with respect to $x$:

$$x^2+y^2=25$$

If is not possible to rearrange this to form a function of $y$ in terms of $x$, so we proceed with implicity differentiation. This is done by differentiating each term in turn.

$$\frac{d}{dx}(x^2)+\frac{d}{dx}(y^2)=\frac{d}{dx}(25)$$

The first and third terms can be done using our knowledge from explicit differentiations:

$$\frac{d}{dx}(x^2)=2x$$

and

$$\frac{d}{dx}(25)=0$$

The second terms requires our knowledge of the chain rule for differentiation:

$$\frac{d}{dx}(y^2)=\frac{d\left(y^2\right)}{dy}\frac{dy}{dx}=2y\frac{dy}{dx}$$

So, bring this all together:

$$2x+2y\frac{dy}{dx}=0$$

or

$$\frac{dy}{dx}=-\frac{x}{y}$$

Worked Example 2:

Differentiate the following with respect to $x$:

$$y^2=x$$

Taking the derivative of both sides of the equation:

$$\frac{d}{dx}(y^2)=\frac{d}{dx}(x)$$ $$\frac{d\left(y^2\right)}{dy}\frac{dy}{dx}=1$$ $$2y\frac{dy}{dx}=1$$ $$\frac{dy}{dx}=\frac{1}{2y}$$

This question is sightly different than before because we could rewrite it thus:

$$y=\sqrt{x}$$

Having only taken the positive root and the derivate of this is:

$$\frac{dy}{dx}=\frac{1}{2x^{\mathrm{½}}}$$

This is the same as the derivative using implicit differentiation.

Worked Example 3:

$$y^3-3x^4{y}^2=25$$

Worked Example 4:

$$\frac{x^2-\sqrt{y}}{x+y}=x$$

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Further Reading:

Implicit Differentiation
(care of www.mathsisfun.com)

Implicity Differentiation Review
(care of www.khanacademy.org)

Implicit Differentiation
(care of math.libretexts.org)

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Questions:

Find $\frac{dy}{dx}$ in terms of $x$,

$$x^3\cot y=2$$

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Solution

Using the product rule for differentiation on the lefthand side implicity,

$$3x^2\cot y-x^3{\mathrm{cosec}}^{2}y\frac{dy}{dx}=0$$ $$\frac{dy}{dx}=\frac{3x^2\cot y}{x^3{\mathrm{cosec}}^{2}y}$$ $$\frac{dy}{dx}=\frac{3\cot y}{x{\mathrm{cosec}}^{2}y}$$

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For the numerator of our differential equation, from the question,

$$\cot y=\frac{2}{x^3}$$ $$3\cot y=\frac{6}{x^3}$$

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For the demoninator,

$${\mathrm{cosec}}^{2}y={\cot }^{2}y+1$$ $${\mathrm{cosec}}^{2}y={\left(\frac{2}{x^3}\right)}^2+1$$ $$x{\mathrm{cosec}}^{2}y=\frac{4}{x^5}+x$$ $$x{\mathrm{cosec}}^{2}y=\frac{4+x^6}{x^5}$$

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Substituting these into $\frac{dy}{dx}$,

$$\frac{dy}{dx}=\frac{6}{x^3}÷\frac{4+x^6}{x^5}$$ $$\frac{dy}{dx}=\frac{6}{x^3}\times \frac{x^5}{4+x^6}$$ $$\frac{dy}{dx}=\frac{6x^2}{4+x^6}$$

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