Implicit Differentiation

Explicit vs. implicit

A function given as $y = f(x)$ is explicit. When $x$ and $y$ are tangled together in one equation instead, differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ (so every $y$ term picks up a $\frac{dy}{dx}$ factor via the chain rule), then solve for $\frac{dy}{dx}$ .

Worked example

Find $\dfrac{dy}{dx}$ if $2x + 3y = \sin x$ .

Solution: Differentiate both sides with respect to $x$ :

2 + 3\dfrac{dy}{dx} = \cos x \;\Rightarrow\; \dfrac{dy}{dx} = \dfrac{\cos x - 2}{3}

Worked example: an inverse trig identity

If $y = \cos^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right)$ , find $\dfrac{dy}{dx}$ .

Solution: Substituting $x = \tan\theta$ turns the fraction into the double-angle identity $\dfrac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta$ , so:

y = \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1}x

Differentiating this simpler explicit form:

\dfrac{dy}{dx} = \dfrac{2}{1+x^2}

Implicit Differentiation

Explicit vs. implicit

Worked example

Find $\dfrac{dy}{dx}$ if $2x + 3y = \sin x$ .

Solution: Differentiate both sides with respect to $x$ :

2 + 3\dfrac{dy}{dx} = \cos x \;\Rightarrow\; \dfrac{dy}{dx} = \dfrac{\cos x - 2}{3}

Worked example: an inverse trig identity

If $y = \cos^{-1}\!\left(\dfrac{1-x^2}{1+x^2}\right)$ , find $\dfrac{dy}{dx}$ .

Solution: Substituting $x = \tan\theta$ turns the fraction into the double-angle identity $\dfrac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta$ , so:

y = \cos^{-1}(\cos 2\theta) = 2\theta = 2\tan^{-1}x

Differentiating this simpler explicit form:

\dfrac{dy}{dx} = \dfrac{2}{1+x^2}

Implicit Differentiation

Explicit vs. implicit

Worked example

Worked example: an inverse trig identity

Continue learning

Implicit Differentiation

Explicit vs. implicit

Worked example

Worked example: an inverse trig identity

Continue learning