Calculus Involving Inverse Trig (Edexcel A-Level Further Mathematics): Revision Notes

📚 Revision Notes
Revision NotesModel AnswersQuizzesFlashcardsExam Questions

5.2.4 Calculus Involving Inverse Trig

Differentiating Inverse Trigonometric Functions

Inverse trigonometric functions, such as arcsin⁡x\arcsin x, arccos⁡x\arccos x, and arctan⁡x\arctan x, are commonly used in calculus. Differentiating these functions requires understanding their derivatives, which can be derived using implicit differentiation or directly memorised as formulas.

Key Differentiation Rules

Derivative of arcsin⁡\arcsin xx:

ddx(arcsin⁡x)=11−x2,∣x∣≤1\frac{d}{dx} (\arcsin x) = \frac{1}{\sqrt{1-x^2}}, \quad |x| \leq 1

Derivative of arccos⁡\arccos xx:

ddx(arccos⁡x)=−11−x2,∣x∣≤1\frac{d}{dx} (\arccos x) = -\frac{1}{\sqrt{1-x^2}}, \quad |x| \leq 1

Derivative of arctan⁡\arctan xx:

ddx(arctan⁡x)=11+x2,x∈R\frac{d}{dx} (\arctan x) = \frac{1}{1+x^2}, \quad x \in \mathbb{R}

Derivative of arccot x\text{arccot}\ x:

ddx(arccot x)=−11+x2,x∈R\frac{d}{dx} (\text{arccot}\ x) = -\frac{1}{1+x^2}, \quad x \in \mathbb{R}

Derivative of arcsec x\text{arcsec}\ x:

ddx(arcsec x)=1∣x∣x2−1,∣x∣>1\frac{d}{dx} (\text{arcsec}\ x) = \frac{1}{|x| \sqrt{x^2-1}}, \quad |x| > 1

Derivative of arccsc x\text{arccsc}\ x:

ddx(arccsc x)=−1∣x∣x2−1,∣x∣>1\frac{d}{dx} (\text{arccsc}\ x) = -\frac{1}{|x| \sqrt{x^2-1}}, \quad |x| > 1

Worked Examples

lightbulbExample

Example 1: Differentiate f(x)=arcsin⁡x+x1−x2f(x) = \arcsin x + x\sqrt{1-x^2}

Step 1: Differentiate term by term:

First term:

ddx(arcsin⁡x)=11−x2\frac{d}{dx} (\arcsin x) = \frac{1}{\sqrt{1-x^2}}

Second term:

Use the product rule for x1−x2x\sqrt{1-x^2}

Let u=xu=x and v=1−x2v = \sqrt{1-x^2}:

ddx(x1−x2)=dudx×v+u×dvdx\frac{d}{dx} (x\sqrt{1-x^2}) = \frac{du}{dx} \times v + u \times \frac{dv}{dx}dudx=1\frac{du}{dx} = 1v=(1−x2)1/2  ⟹  dvdx=−x1−x2v = (1-x^2)^{1/2} \implies \frac{dv}{dx} = \frac{-x}{\sqrt{1-x^2}}

Substitute:

ddx(x1−x2)=1×1−x2+x×−x1−x2\frac{d}{dx} (x\sqrt{1-x^2}) = 1 \times \sqrt{1-x^2} + x \times \frac{-x}{\sqrt{1-x^2}} =1−x2−x21−x2= \sqrt{1-x^2} - \frac{x^2}{\sqrt{1-x^2}}

Simplify:

ddx(x1−x2)=1−x21−x2=1−x2\frac{d}{dx} (x\sqrt{1-x^2}) = \frac{1-x^2}{\sqrt{1-x^2}} = \sqrt{1-x^2}

Step 2: Combine results:

ddx(arcsin⁡x+x1−x2)=11−x2+1−x2\frac{d}{dx} \left( \arcsin x + x\sqrt{1-x^2} \right) = \frac{1}{\sqrt{1-x^2}} + \sqrt{1-x^2}
lightbulbExample

Example 2**:** Differentiate f(x)=12arctan⁡(x2)f(x) = \frac{1}{2} \arctan(x^2)

Step 1: Apply the chain rule:

Let u=x2u = x^2

Then f(x)=12arctan⁡u f(x) = \frac{1}{2} \arctan u and dudx=2x\frac{du}{dx} = 2x

Step 2: Differentiate arctan⁡u\arctan u:

ddx(arctan⁡u)=11+u2×dudx\frac{d}{dx} (\arctan u) = \frac{1}{1+u^2} \times \frac{du}{dx}

Step 3: Substitute back:

ddx(12arctan⁡(x2))=12×11+(x2)2×2x\frac{d}{dx} \left( \frac{1}{2} \arctan(x^2) \right) = \frac{1}{2} \times \frac{1}{1+(x^2)^2} \times 2x =x1+x4= \frac{x}{1+x^4}

Result:

ddx(12arctan⁡(x2))=x1+x4\frac{d}{dx} \left( \frac{1}{2} \arctan(x^2) \right) = \frac{x}{1+x^4}
lightbulbExample

Example 3: Differentiate f(x)=xarccos⁡xf(x) = x \arccos x

Step 1: Use the product rule:

Let u=xu = x and v=arccos⁡xv = \arccos x

ddx(xarccos⁡x)=dudx×v+u×dvdx\frac{d}{dx} (x \arccos x) = \frac{du}{dx} \times v + u \times \frac{dv}{dx}

Step 2: Differentiate uu and vv:

dudx=1\frac{du}{dx} = 1dvdx=−11−x2\frac{dv}{dx} = -\frac{1}{\sqrt{1-x^2}}

Step 3: Substitute back:

ddx(xarccos⁡x)=1×arccos⁡x+x×(−11−x2)\frac{d}{dx} (x \arccos x) = 1 \times \arccos x + x \times \left(-\frac{1}{\sqrt{1-x^2}}\right)

Step 4: Simplify:

ddx(xarccos⁡x)=arccos⁡x−x1−x2\frac{d}{dx} (x \arccos x) = \arccos x - \frac{x}{\sqrt{1-x^2}}

Note Summary

infoNote

Common Mistakes:

  1. Mixing up derivatives of arcsin⁡x\arcsin x and arccos⁡x\arccos x: Remember arcsin⁡x\arcsin x has a positive derivative, whereas arccos⁡x\arccos x has a negative one.

  2. Forgetting the chain rule: When differentiating compositions like arctan⁡(x2)\arctan(x^2), always apply the chain rule.

  3. Mismanaging square roots: Be careful with expressions like 1−x2\sqrt{1-x^2}, especially when simplifying.

  1. Ignoring domain restrictions: Inverse trigonometric functions have specific domains,

📑e.g., ∣x∣≤1|x| \leq 1 for arcsin⁡x\arcsin x and arccos⁡x\arccos x.

  1. Omitting the product rule: Don't forget to apply the product rule for terms like xarccos⁡xx \arccos x
infoNote

Key Formulas:

  1. Basic Derivatives:
ddx(arcsin⁡x)=11−x2,ddx(arccos⁡x)=−11−x2\frac{d}{dx} (\arcsin x) = \frac{1}{\sqrt{1-x^2}}, \quad \frac{d}{dx} (\arccos x) = -\frac{1}{\sqrt{1-x^2}}ddx(arctan⁡x)=11+x2,ddx(arccot x)=−11+x2\frac{d}{dx} (\arctan x) = \frac{1}{1+x^2}, \quad \frac{d}{dx} (\text{arccot}\ x) = -\frac{1}{1+x^2}ddx(arcsec x)=1∣x∣x2−1,ddx(arccsc x)=−1∣x∣x2−1\frac{d}{dx} (\text{arcsec}\ x) = \frac{1}{|x| \sqrt{x^2-1}}, \quad \frac{d}{dx} (\text{arccsc}\ x) = -\frac{1}{|x| \sqrt{x^2-1}}
  1. Product Rule:
ddx(u×v)=dudx×v+u×dvdx\frac{d}{dx} (u \times v) = \frac{du}{dx} \times v + u \times \frac{dv}{dx}
  1. Chain Rule:
ddxf(g(x))=f′(g(x))×g′(x)\frac{d}{dx} f(g(x)) = f'(g(x)) \times g'(x)

Explore Edexcel A-Level Further Mathematics Model Answers by Topics

Volumes of Revolution

Methods in Calculus

Explore Edexcel A-Level Further Mathematics Quizzes by Topics

Volumes of Revolution

Methods in Calculus

Explore Edexcel A-Level Further Mathematics Flashcards by Topics

Volumes of Revolution

Methods in Calculus

Explore Edexcel A-Level Further Mathematics Exam Questions by Topics

Volumes of Revolution

Methods in Calculus

Join 100,000+ A-Level students studying Revision Notes with us.

Select your subjects, and get access to A+ resources today.