Implicit Functions

So far, we have mostly differentiated functions written in the explicit form y=f(x).y = f(x). However, many relations between xx and yy are given in a form where yy is not isolated, for example f(x,y)=0.f(x,y)=0. Such relations are called implicit functions.

Examples include: x2+y2=1,x3+y3=3axy,y+sin⁡y=cos⁡x.x^2+y^2=1, \qquad x^3+y^3=3axy, \qquad y+\sin y = \cos x. In these equations, yy depends on xx, but it is not written directly as a simple formula y=f(x)y=f(x).

Method of differentiating an implicit function

  1. Differentiate both sides of the equation with respect to xx.
  2. Whenever a term involving yy is differentiated, treat yy as a function of xx and apply the chain rule. For example, ddx(y2)=2ydydx,ddx(sin⁡y)=cos⁡ydydx.\frac{d}{dx}(y^2)=2y\frac{dy}{dx}, \qquad \frac{d}{dx}(\sin y)=\cos y\frac{dy}{dx}.
  3. Collect all terms containing dydx\dfrac{dy}{dx} on one side.
  4. Factor out dydx\dfrac{dy}{dx}.
  5. Solve for dydx\dfrac{dy}{dx}.

This method is called implicit differentiation.

Derivatives of Inverse Trigonometric Functions

The derivatives of inverse trigonometric functions are standard and very important. These formulas are usually obtained by implicit differentiation.

For the principal branches, we have:

  • ddx(sin⁡−1x)=11−x2,−1<x<1\frac{d}{dx}(\sin^{-1}x)=\frac{1}{\sqrt{1-x^2}}, \qquad -1<x<1

  • ddx(cos⁡−1x)=−11−x2,−1<x<1\frac{d}{dx}(\cos^{-1}x)=\frac{-1}{\sqrt{1-x^2}}, \qquad -1<x<1

  • ddx(tan⁡−1x)=11+x2,x∈R\frac{d}{dx}(\tan^{-1}x)=\frac{1}{1+x^2}, \qquad x\in\mathbb{R}

  • ddx(cot⁡−1x)=−11+x2,x∈R\frac{d}{dx}(\cot^{-1}x)=\frac{-1}{1+x^2}, \qquad x\in\mathbb{R}

  • ddx(sec⁡−1x)=1∣x∣x2−1,∣x∣>1\frac{d}{dx}(\sec^{-1}x)=\frac{1}{|x|\sqrt{x^2-1}}, \qquad |x|>1

  • ddx(csc⁡−1x)=−1∣x∣x2−1,∣x∣>1\frac{d}{dx}(\csc^{-1}x)=\frac{-1}{|x|\sqrt{x^2-1}}, \qquad |x|>1

These formulas are valid only where the given inverse trigonometric functions are defined and differentiable.

Differentiation by Trigonometric Substitution

Sometimes the direct chain-rule differentiation of an inverse trigonometric expression becomes long and algebraically messy. In such cases, trigonometric substitution is often the smartest method.

Useful substitutions:

  • For expressions involving a2−x2,\sqrt{a^2-x^2}, use x=asin⁡θorx=acos⁡θ.x=a\sin\theta \quad \text{or} \quad x=a\cos\theta.

  • For expressions involving a2+x2ora2+x2,a^2+x^2 \quad \text{or} \quad \sqrt{a^2+x^2}, use x=atan⁡θorx=acot⁡θ.x=a\tan\theta \quad \text{or} \quad x=a\cot\theta.

  • For expressions involving x2−a2,\sqrt{x^2-a^2}, use x=asec⁡θorx=acsc⁡θ.x=a\sec\theta \quad \text{or} \quad x=a\csc\theta.

Also, the following expressions strongly suggest the substitution x=tan⁡θx=\tan\theta because of double-angle identities:

  • 2x1+x2corresponds tosin⁡2θ\frac{2x}{1+x^2} \quad \text{corresponds to} \quad \sin 2\theta

  • 1−x21+x2corresponds tocos⁡2θ\frac{1-x^2}{1+x^2} \quad \text{corresponds to} \quad \cos 2\theta

  • 2x1−x2corresponds totan⁡2θ\frac{2x}{1-x^2} \quad \text{corresponds to} \quad \tan 2\theta

Whenever such substitutions are used, care must be taken with the principal values of inverse trigonometric functions.

Solved Examples

Example 1: Basic Implicit Differentiation

Find dydx\frac{dy}{dx} if x−y=πx-y=\pi.

Solution: Differentiate both sides with respect to xx: ddx(x−y)=ddx(π).\frac{d}{dx}(x-y)=\frac{d}{dx}(\pi). Since the derivative of xx is 11, the derivative of yy with respect to xx is dydx\dfrac{dy}{dx}, and the derivative of the constant π\pi is 00, we get 1−dydx=0.1-\frac{dy}{dx}=0. Therefore, dydx=1.\frac{dy}{dx}=1.

Answer: 11

Example 2: Implicit Differentiation with Chain Rule

Find dydx\frac{dy}{dx} if y+sin⁡y=cos⁡xy+\sin y=\cos x.

Solution: Differentiate both sides with respect to xx: ddx(y)+ddx(sin⁡y)=ddx(cos⁡x).\frac{d}{dx}(y)+\frac{d}{dx}(\sin y)=\frac{d}{dx}(\cos x). Now, ddx(y)=dydx,\frac{d}{dx}(y)=\frac{dy}{dx}, and by the chain rule, ddx(sin⁡y)=cos⁡y⋅dydx.\frac{d}{dx}(\sin y)=\cos y\cdot\frac{dy}{dx}. Also, ddx(cos⁡x)=−sin⁡x.\frac{d}{dx}(\cos x)=-\sin x. So, dydx+cos⁡ydydx=−sin⁡x.\frac{dy}{dx}+\cos y\frac{dy}{dx}=-\sin x. Factor out dydx\dfrac{dy}{dx}: dydx(1+cos⁡y)=−sin⁡x.\frac{dy}{dx}(1+\cos y)=-\sin x. Hence, dydx=−sin⁡x1+cos⁡y,\frac{dy}{dx}=\frac{-\sin x}{1+\cos y}, provided 1+cos⁡y≠01+\cos y\ne 0.

Answer: −sin⁡x1+cos⁡y\frac{-\sin x}{1+\cos y}

Example 3: Implicit Differentiation with Product Rule

Find dydx\frac{dy}{dx} if x2+xy+y2=100x^2+xy+y^2=100.

Solution: Differentiate both sides with respect to xx: ddx(x2)+ddx(xy)+ddx(y2)=ddx(100).\frac{d}{dx}(x^2)+\frac{d}{dx}(xy)+\frac{d}{dx}(y^2)=\frac{d}{dx}(100). Now, ddx(x2)=2x,\frac{d}{dx}(x^2)=2x, ddx(xy)=xdydx+y\frac{d}{dx}(xy)=x\frac{dy}{dx}+y by the product rule, and ddx(y2)=2ydydx\frac{d}{dx}(y^2)=2y\frac{dy}{dx} by the chain rule. So, 2x+xdydx+y+2ydydx=0.2x + x\frac{dy}{dx} + y + 2y\frac{dy}{dx} = 0. Group the derivative terms: xdydx+2ydydx=−(2x+y).x\frac{dy}{dx}+2y\frac{dy}{dx}=-(2x+y). Factor out dydx\dfrac{dy}{dx}: dydx(x+2y)=−(2x+y).\frac{dy}{dx}(x+2y)=-(2x+y). Therefore, dydx=−(2x+y)x+2y,\frac{dy}{dx}=\frac{-(2x+y)}{x+2y}, provided x+2y≠0x+2y\ne 0.

Answer: −(2x+y)x+2y\frac{-(2x+y)}{x+2y}

Example 4: Complex Implicit Differentiation

Find dydx\frac{dy}{dx} if sin⁡2y+cos⁡(xy)=k\sin^2 y + \cos(xy)=k, where kk is a constant.

Solution: Differentiate both sides with respect to xx: ddx(sin⁡2y)+ddx(cos⁡(xy))=ddx(k).\frac{d}{dx}(\sin^2 y)+\frac{d}{dx}(\cos(xy))=\frac{d}{dx}(k). Since kk is constant, its derivative is 00.

For the first term, ddx(sin⁡2y)=2sin⁡ycos⁡ydydx=sin⁡2ydydx.\frac{d}{dx}(\sin^2 y)=2\sin y\cos y\frac{dy}{dx}=\sin 2y\frac{dy}{dx}.

For the second term, use the chain rule together with the product rule: ddx(cos⁡(xy))=−sin⁡(xy)⋅ddx(xy),\frac{d}{dx}(\cos(xy))=-\sin(xy)\cdot\frac{d}{dx}(xy), where ddx(xy)=xdydx+y.\frac{d}{dx}(xy)=x\frac{dy}{dx}+y. Thus, sin⁡2ydydx−sin⁡(xy)(xdydx+y)=0.\sin 2y\frac{dy}{dx} - \sin(xy)(x\frac{dy}{dx}+y)=0. Expand: sin⁡2ydydx−xsin⁡(xy)dydx−ysin⁡(xy)=0.\sin 2y\frac{dy}{dx} - x\sin(xy)\frac{dy}{dx} - y\sin(xy)=0. Group the terms containing dydx\dfrac{dy}{dx}: dydx(sin⁡2y−xsin⁡(xy))=ysin⁡(xy).\frac{dy}{dx}\big(\sin 2y - x\sin(xy)\big)=y\sin(xy). Hence, dydx=ysin⁡(xy)sin⁡2y−xsin⁡(xy).\frac{dy}{dx}=\frac{y\sin(xy)}{\sin 2y - x\sin(xy)}.

Answer: ysin⁡(xy)sin⁡2y−xsin⁡(xy)\frac{y\sin(xy)}{\sin 2y - x\sin(xy)}

Example 5: Derivative of an Inverse Trigonometric Function using Chain Rule

Find the derivative of y=tan⁡−1(3x)y=\tan^{-1}(3x).

Solution: Use the standard formula ddx(tan⁡−1u)=11+u2⋅dudx.\frac{d}{dx}(\tan^{-1}u)=\frac{1}{1+u^2}\cdot\frac{du}{dx}. Here, u=3x.u=3x. So, dudx=3.\frac{du}{dx}=3. Therefore, dydx=11+(3x)2⋅3=31+9x2.\frac{dy}{dx}=\frac{1}{1+(3x)^2}\cdot 3 = \frac{3}{1+9x^2}.

Answer: 31+9x2\frac{3}{1+9x^2}

Example 6: Trigonometric Substitution (Sine)

Find dydx\frac{dy}{dx} if y=sin⁡−1(2x1+x2)y=\sin^{-1}\left(\frac{2x}{1+x^2}\right) for ∣x∣<1|x|<1.

Solution: The expression 2x1+x2\frac{2x}{1+x^2} suggests the identity sin⁡2θ=2tan⁡θ1+tan⁡2θ.\sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta}. Let x=tan⁡θ.x=\tan\theta. Then θ=tan⁡−1x.\theta=\tan^{-1}x. Substituting, y=sin⁡−1(2tan⁡θ1+tan⁡2θ)=sin⁡−1(sin⁡2θ).y=\sin^{-1}\left(\frac{2\tan\theta}{1+\tan^2\theta}\right)=\sin^{-1}(\sin 2\theta). For ∣x∣<1|x|<1, we have θ∈(−π4,π4)\theta\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right), so 2θ∈(−π2,π2),2\theta\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right), which lies in the principal range of sin⁡−1\sin^{-1}. Hence, y=2θ=2tan⁡−1x.y=2\theta=2\tan^{-1}x. Now differentiate: dydx=2⋅11+x2=21+x2.\frac{dy}{dx}=2\cdot\frac{1}{1+x^2}=\frac{2}{1+x^2}.

Answer: 21+x2\frac{2}{1+x^2}

Example 7: Trigonometric Substitution (Cosine)

Find dydx\frac{dy}{dx} if y=cos⁡−1(1−x21+x2), 0<x<1y=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right),\ 0<x<1.

Solution: The expression 1−x21+x2\frac{1-x^2}{1+x^2} suggests the identity cos⁡2θ=1−tan⁡2θ1+tan⁡2θ.\cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta}. Let x=tan⁡θ.x=\tan\theta. Since 0<x<10<x<1, we get 0<θ<π4,0<\theta<\frac{\pi}{4}, so 0<2θ<π2.0<2\theta<\frac{\pi}{2}. Now, y=cos⁡−1(1−tan⁡2θ1+tan⁡2θ)=cos⁡−1(cos⁡2θ).y=\cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right)=\cos^{-1}(\cos 2\theta). Because 2θ2\theta lies in the principal range [0,π][0,\pi] of cos⁡−1\cos^{-1}, we may write y=2θ=2tan⁡−1x.y=2\theta=2\tan^{-1}x. Differentiate: dydx=2⋅11+x2=21+x2.\frac{dy}{dx}=2\cdot\frac{1}{1+x^2}=\frac{2}{1+x^2}.

Answer: 21+x2\frac{2}{1+x^2}

Example 8: Trigonometric Substitution (Tangent)

Find dydx\frac{dy}{dx} if y=tan⁡−1(3x−x31−3x2), −13<x<13y=\tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right),\ -\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}.

Solution: The inner expression matches the identity tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ.\tan 3\theta = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta}. Let x=tan⁡θ.x=\tan\theta. Then θ=tan⁡−1x.\theta=\tan^{-1}x. Substituting, y=tan⁡−1(tan⁡3θ).y=\tan^{-1}(\tan 3\theta). Because −13<x<13,-\frac{1}{\sqrt{3}}<x<\frac{1}{\sqrt{3}}, we have −π6<θ<π6,-\frac{\pi}{6}<\theta<\frac{\pi}{6}, so −π2<3θ<π2.-\frac{\pi}{2}<3\theta<\frac{\pi}{2}. Thus 3θ3\theta lies in the principal range of tan⁡−1\tan^{-1}, and therefore y=3θ=3tan⁡−1x.y=3\theta=3\tan^{-1}x. Differentiating, dydx=3⋅11+x2=31+x2.\frac{dy}{dx}=3\cdot\frac{1}{1+x^2}=\frac{3}{1+x^2}.

Answer: 31+x2\frac{3}{1+x^2}

Example 9: Complementary Angles in Inverse Trig

Find dydx\frac{dy}{dx} if y=sin⁡−1(1−x21+x2), 0<x<1y=\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right),\ 0<x<1.

Solution: Using the same substitution as before, let x=tan⁡θ,0<θ<π4.x=\tan\theta, \qquad 0<\theta<\frac{\pi}{4}. Then 1−x21+x2=cos⁡2θ.\frac{1-x^2}{1+x^2}=\cos 2\theta. So, y=sin⁡−1(cos⁡2θ).y=\sin^{-1}(\cos 2\theta). Now use the identity cos⁡A=sin⁡(π2−A).\cos A = \sin\left(\frac{\pi}{2}-A\right). Hence, y=sin⁡−1(sin⁡(π2−2θ)).y=\sin^{-1}\left(\sin\left(\frac{\pi}{2}-2\theta\right)\right). Since 0<θ<π40<\theta<\frac{\pi}{4}, we get 0<π2−2θ<π2,0<\frac{\pi}{2}-2\theta<\frac{\pi}{2}, which lies in the principal range of sin⁡−1\sin^{-1}. Therefore, y=\frac{\pi}{2}-2\theta= rac{\pi}{2}-2\tan^{-1}x. Now differentiate: dydx=0−2⋅11+x2=−21+x2.\frac{dy}{dx}=0-2\cdot\frac{1}{1+x^2}=\frac{-2}{1+x^2}.

Answer: −21+x2\frac{-2}{1+x^2}

Example 10: Inverse Secant Substitution

Find dydx\frac{dy}{dx} if y=sec⁡−1(12x2−1)y=\sec^{-1}\left(\frac{1}{2x^2-1}\right) for 0<x<10<x<1.

Solution: The denominator 2x2−12x^2-1 suggests the identity cos⁡2θ=2cos⁡2θ−1.\cos 2\theta = 2\cos^2\theta - 1. Let x=cos⁡θ.x=\cos\theta. Since 0<x<10<x<1, we may take 0<θ<π2.0<\theta<\frac{\pi}{2}. Then 2x2−1=2cos⁡2θ−1=cos⁡2θ.2x^2-1 = 2\cos^2\theta - 1 = \cos 2\theta. Hence, 12x2−1=1cos⁡2θ=sec⁡2θ.\frac{1}{2x^2-1} = \frac{1}{\cos 2\theta} = \sec 2\theta. So, y=sec⁡−1(sec⁡2θ).y = \sec^{-1}(\sec 2\theta). On the principal branch, this simplifies to y=2θ=2cos⁡−1xy=2\theta=2\cos^{-1}x for the appropriate interval.

Now differentiate: dydx=2⋅ddx(cos⁡−1x)=2(−11−x2)=−21−x2.\frac{dy}{dx}=2\cdot\frac{d}{dx}(\cos^{-1}x)=2\left(\frac{-1}{\sqrt{1-x^2}}\right)=\frac{-2}{\sqrt{1-x^2}}.

Answer: −21−x2\frac{-2}{\sqrt{1-x^2}}