Continuity and Differentiability · Lesson 3 of 8
Implicit and Inverse Trigonometric Differentiation
“Differentiate relationships that do not isolate y, then use that method to understand inverse trigonometric derivatives and their domains.”
• Distinguish explicit formulas from implicit relationships between x and y. • Differentiate expressions involving y by including the required dy/dx factors. • Derive inverse sine differentiation using its principal range. • Use inverse trigonometric derivative formulas on their appropriate domains. • Simplify inverse trigonometric compositions only after checking principal values. • Solve implicit and composite derivative problems with clear denominator restrictions.
When a relationship does not isolate y
An explicit function gives y directly in terms of x, as in y=x². An equation such as x²+xy+y²=100 mixes the two variables. Solving it for y first may be inconvenient, and it may describe more than one branch. Implicit differentiation lets us work directly with the relationship on a branch where y depends differentiably on x.
An explicit description has the form y=f(x). An implicit description gives a relationship involving x and y without isolating y. When differentiating such a relationship with respect to x, treat y as a function of x on the branch being considered.
The chain rule explains what happens to terms involving y. The derivative of y² with respect to x is 2y·dy/dx, not merely 2y. The derivative of sin y is cos y·dy/dx. For a product xy, both x and y change, so the derivative is y+x·dy/dx. After differentiating, collect the terms containing dy/dx and solve for it if the coefficient is nonzero.
Problem
Find dy/dx if x−y=π.
- 1.An explicit approach gives y=x−π, hence dy/dx=1 because π is a constant.
- 2.For a direct approach, differentiate the given equation with respect to x: 1−dy/dx=0.
- 3.Rearrange to obtain dy/dx=1 again. The two approaches agree because they describe the same function.
Problem
Find dy/dx if y+sin y=cos x.
- 1.Differentiate each side with respect to x. The derivative of y is y′; the derivative of sin y is cos y·y′; the derivative of cos x is −sin x.
- 2.Thus y′+cos y·y′=−sin x. Factor y′ to obtain (1+cos y)y′=−sin x.
- 3.Where 1+cos y≠0, divide to get y′=−sin x/(1+cos y).
- 4.The excluded coefficient values are y=(2n+1)π. Division by zero would not establish a derivative at such a point; the displayed calculation applies where the coefficient is nonzero.
Problem
Find dy/dx for x²+xy+y²=100.
- 1.Differentiate the terms separately: d(x²)/dx=2x, d(xy)/dx=y+xy′, and d(y²)/dx=2yy′.
- 2.The constant on the right differentiates to zero, giving 2x+y+xy′+2yy′=0.
- 3.Group the derivative terms: (x+2y)y′=−(2x+y).
- 4.Therefore y′=−(2x+y)/(x+2y) wherever x+2y≠0. This expression can depend on both x and y because both coordinates identify a point on the implicit curve.
Problem
Differentiate sin²y+cos(xy)=κ, where κ is a constant.
- 1.The first term is [sin y]², so its derivative is 2sin y cos y·y′.
- 2.For cos(xy), first differentiate the outer cosine and then the product xy: its derivative is −sin(xy)(y+xy′).
- 3.The differentiated equation is 2sin y cos y·y′−y sin(xy)−x sin(xy)y′=0.
- 4.Collecting y′ gives [2sin y cos y−x sin(xy)]y′=y sin(xy).
- 5.Divide by the bracket when it is nonzero. Thus y′=y sin(xy)/[2sin y cos y−x sin(xy)]. A missing product-rule term would change both the numerator and denominator.
While differentiating with respect to x, y still changes when it is a function of x. Replacing d(y²)/dx by 2y, or d(xy)/dx by y alone, loses the dependence on x. Write y′ explicitly until all such terms have been collected.
Inverse trigonometric functions: the range controls the sign
The notation sin⁻¹x means the inverse sine, also written arcsin x; it does not mean 1/sin x. Inverse trigonometric functions choose a principal value from a specified range. Those range choices decide the signs of square roots that appear during differentiation.
The inverse sine is continuous on [−1,1], but its derivative formula applies on the open interval (−1,1). To find the derivative, write y=sin⁻¹x, so x=sin y. This is an implicit relationship of the type we have just learned to differentiate. On the principal range −π/2≤y≤π/2, cosine is nonnegative, and it is positive away from the endpoints.
Problem
Find dy/dx for y=sin⁻¹x on −1<x<1.
- 1.Rewrite the equation as x=sin y. Differentiating with respect to x gives 1=cos y·y′.
- 2.For −1<x<1, the principal angle y lies strictly between −π/2 and π/2. Therefore cos y>0 and y′=1/cos y.
- 3.Using sin y=x, we obtain cos²y=1−x². The positive cosine on this range gives cos y=√(1−x²).
- 4.Thus y′=1/√(1−x²). At x=±1 the denominator vanishes; these endpoints belong to the function domain but not to this finite derivative domain.
For y=cos⁻¹x, write x=cos y with 0≤y≤π. On the interior of this range sin y is positive. Differentiating gives 1=−sin y·y′, hence the negative inverse-cosine formula. For y=tan⁻¹x, write x=tan y. Then 1=sec²y·y′ and sec²y=1+x², giving the inverse-tangent formula.
| Function | Principal range | Function domain | Derivative domain |
|---|---|---|---|
| sin⁻¹x | [−π/2,π/2] | [−1,1] | (−1,1) |
| cos⁻¹x | [0,π] | [−1,1] | (−1,1) |
| tan⁻¹x | (−π/2,π/2) | All real numbers | All real numbers |
Inverse sine, inverse cosine, and inverse tangent are continuous on their full respective domains. Continuity at the endpoints of the first two does not make their derivative formulas finite there. We use the continuity of inverse trigonometric functions here without a separate proof, as in the source.
For a differentiable inner function u(x), multiply each standard derivative by u′(x). For example, d[sin⁻¹u]/dx=u′/√(1−u²) where |u|<1. The outer derivative is evaluated at u, so both the denominator and its domain depend on the whole inner expression. A related formula needed in mixed work follows from cot⁻¹u=π/2−tan⁻¹u, using the principal range (0,π): its derivative is −u′/(1+u²).
Why simplifying an inverse composition needs a range check
A trigonometric identity may turn an argument into sin(2θ) or tan(3θ). An inverse function returns the principal angle with that trigonometric value; it does not undo every possible angle. Simplification can make differentiation easier, provided we first show that the angle lies in the appropriate principal range.
Problem
Differentiate y=tan⁻¹[(3x−x³)/(1−3x²)] for −1/√3<x<1/√3.
- 1.Set θ=tan⁻¹x. The interval for x implies −π/6<θ<π/6, so −π/2<3θ<π/2.
- 2.The tangent triple-angle identity makes the inner fraction tan(3θ). The denominator 1−3x² is positive on the given interval.
- 3.Because 3θ lies in the principal range of inverse tangent, y=tan⁻¹(tan3θ)=3θ=3tan⁻¹x.
- 4.Differentiate the simpler expression to get y′=3/(1+x²). The interval check justified the simplification before it was used.
Problem
Differentiate y=sin⁻¹[2x/(1+x²)] on its differentiable intervals.
- 1.The inner value always lies in [−1,1] because |2x|≤1+x². Thus the function is defined for all real x.
- 2.Differentiate the inner quotient: u′=2(1−x²)/(1+x²)². Also 1−u²=(1−x²)²/(1+x²)².
- 3.Taking a square root gives √(1−u²)=|1−x²|/(1+x²), not (1−x²)/(1+x²) without an absolute value.
- 4.For x≠±1, the chain rule therefore gives y′=2(1−x²)/[(1+x²)|1−x²|]. It is 2/(1+x²) when |x|<1 and −2/(1+x²) when |x|>1.
- 5.At x=1 the left and right slopes are 1 and −1; at x=−1 they are −1 and 1. These corners have no two-sided derivative. This is why the answer cannot be 2/(1+x²) for every real input.
Problem
Simplify and differentiate three inverse trigonometric expressions on the indicated intervals.
- 1.For y=cos⁻¹[(1−x²)/(1+x²)], 0<x<1, set θ=tan⁻¹x. Then 0<2θ<π/2 and the argument is cos2θ. Hence y=2θ and y′=2/(1+x²).
- 2.For y=sin⁻¹[(1−x²)/(1+x²)], 0<x<1, the argument is sin(π/2−2θ). This angle lies between 0 and π/2, so y=π/2−2θ and y′=−2/(1+x²).
- 3.For y=cos⁻¹[2x/(1+x²)], −1<x<1, again set θ=tan⁻¹x. The argument is cos(π/2−2θ). Now π/2−2θ lies between 0 and π, the inverse-cosine range.
- 4.Thus y=π/2−2θ and y′=−2/(1+x²). In each case the same rational identities are used, but a different outer inverse function requires a different range check.
Problem
Differentiate y=sin⁻¹[2x√(1−x²)] for −1/√2<x<1/√2.
- 1.Put θ=sin⁻¹x. The stated interval gives −π/4<θ<π/4, so −π/2<2θ<π/2.
- 2.On the principal sine range, cos θ=√(1−x²). Therefore 2x√(1−x²)=2sin θ cos θ=sin2θ.
- 3.The angle 2θ is in the inverse-sine range, so y=2θ=2sin⁻¹x.
- 4.Differentiate to obtain y′=2/√(1−x²). Without the interval restriction, inverse sine could return a reflected angle instead of 2θ.
Problem
Find dy/dx for y=sec⁻¹[1/(2x²−1)], where 0<x<1/√2.
- 1.The inner value is less than −1. The principal inverse-secant angle is therefore in (π/2,π). Taking the reciprocal of sec y gives cos y=2x²−1.
- 2.Differentiate the latter equation: −sin y·y′=4x.
- 3.On (π/2,π), sin y is positive. From cos y=2x²−1, sin y=√[1−(2x²−1)²]=2x√(1−x²), using x>0.
- 4.Thus y′=−4x/[2x√(1−x²)]=−2/√(1−x²). The implicit method obtains the derivative with the correct sign without memorising an additional general formula.
The identity √(A²)=|A| prevents a negative square root. Losing that absolute value can erase an interval-dependent sign in inverse trigonometric derivatives. Similarly, sin⁻¹(sin θ)=θ and tan⁻¹(tan θ)=θ require θ to lie in the chosen principal range.
Check your understanding
The questions below combine basic implicit calculations with the range decisions behind inverse derivatives. When two answers differ only by a sign, look for a chain-rule sign or a principal-range choice before doing more algebra.
Quiz
If 2x+3y=sin x, what is y′?
If y+sin y=cos x, which expression gives y′ when its denominator is nonzero?
What is d(cos⁻¹x)/dx for −1<x<1?
Where does sin⁻¹x have the finite derivative 1/√(1−x²)?
For x²+xy+y²=100, which is y′ when x+2y≠0?
On −1/√3<x<1/√3, what is the derivative of tan⁻¹[(3x−x³)/(1−3x²)]?
For f(x)=sin⁻¹[2x/(1+x²)], what is f′(2)?
For y=sec⁻¹[1/(2x²−1)] on 0<x<1/√2, what is y′?
Practice Problems
- Find y′ for 2x+3y=sin x, 2x+3y=sin y, and ax+by²=cos y. State the coefficients you divide by.
- Differentiate xy+y²=tan x+y. Collect all terms involving y′ before solving.
- Find y′ for x³+x²y+xy²+y³=81. Show the product rule for each mixed term.
- Differentiate sin²y+cos(xy)=κ and sin²x+cos²y=1. Identify where the resulting derivative formulas have nonzero denominators.
- Derive the formula for d(cos⁻¹x)/dx by writing x=cos y and using the principal range of y.
- For sin⁻¹[2x/(1+x²)], find the function domain and derivative on each of the intervals separated by −1 and 1. Explain what happens at the two boundary inputs.
- Use a principal-angle substitution to differentiate tan⁻¹[(3x−x³)/(1−3x²)] on −1/√3<x<1/√3.
- For 0<x<1, differentiate cos⁻¹[(1−x²)/(1+x²)] and sin⁻¹[(1−x²)/(1+x²)]. Explain why the signs differ.
- Differentiate cos⁻¹[2x/(1+x²)] for −1<x<1. Check the angle that the inverse cosine returns.
- Differentiate sin⁻¹[2x√(1−x²)] on −1/√2<x<1/√2. Explain the purpose of the stated interval.
- Find the derivative of sec⁻¹[1/(2x²−1)] for 0<x<1/√2 by first finding cos y.
- A student writes d[sin(y²)]/dx=cos(y²)·2y. Identify the missing factor, and explain how it follows from viewing y as a function of x.
Key Takeaways
• Implicit differentiation treats y as a function of x and uses the chain rule for every expression involving y. • Collect y′ terms and check that their coefficient is nonzero before dividing. • Inverse trigonometric derivatives follow from implicit equations and the principal ranges of the inverse functions. • A function’s domain and its finite derivative domain need not be identical. • An inverse function returns a principal angle, so inverse-composition simplifications require interval checks. • Absolute values arising from square roots of squares preserve the correct signs and derivative intervals.