∫ SJ Maths Mathematics Notes
UP PGT Tracker UP TGT Tracker
Calculus

Differentiation of trigonometric functions

The differentiation of trigonometric functions forms the bedrock of differential calculus, establishing standard rate-of-change formulas for sine, cosine, tangent, and their reciprocals.

For: UP PGT Mathematics UP TGT Mathematics DSSSB PGT / TGT KVS / NVS PGT LT Grade
📖 Topic progress
○ Mark as complete

Definition and statement

First Principles & Standard Derivatives of Trigonometric Functions
Let $f(x) = \sin x$. Then $f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h} = \cos x$ for all $x \in \mathbb{R}$. Similarly, $\frac{d}{dx}(\cos x) = -\sin x$, $\frac{d}{dx}(\tan x) = \sec^2 x$, $\frac{d}{dx}(\cot x) = -\csc^2 x$, $\frac{d}{dx}(\sec x) = \sec x \tan x$, and $\frac{d}{dx}(\csc x) = -\csc x \cot x$.

📐 Geometric Interpretation

The derivative represents the instantaneous slope of the tangent line to the trigonometric curve at any point $x$. For instance, the slope of $y = \sin x$ at $x = 0$ is $\cos(0) = 1$, matching the slope of the line $y = x$ near the origin.

⚙️ Algebraic & Structural Meaning

These formulas provide the analytical machinery required to solve higher-order differential equations, optimization problems involving periodic phenomena, and Taylor series expansions of transcendental functions.

Conditions and hypotheses

Check each condition and what changes when it is omitted:

Hypothesis / Condition Domain / Interval Necessity What Fails If Condition Dropped
Continuity of the Trigonometric Function \mathbb{R} \text{ (or respective domains)} Strictly Mandatory / Necessary At points of discontinuity (e.g., $x = \frac{\pi}{2} + k\pi$ for $\tan x$), the derivative does not exist as the function is undefined.
Differentiability on the Open Domain Open intervals within domain Strictly Mandatory Failure to verify domain restrictions leads to invalid evaluation of derivatives involving inverse or reciprocal trigonometric functions.
Limit Existence for First Principles Neighborhood of $x$ Essential for derivative existence If the limit of incremental ratios fails to converge, the derivative is undefined at that point.

Useful results and shortcuts

Use these identities and results when their conditions apply:

01

Chain Rule for Powers of Sine/Cosine

$$\frac{d}{dx}[\sin^n x] = n \sin^{n-1} x \cos x$$

Rule: Instantly bring down the power, reduce the exponent by 1, and multiply by the derivative of the inner function.

Application: Directly used in UP PGT/TGT questions asking for higher-order derivatives or tangents of power functions.
02

Logarithmic Differentiation for Product/Quotient Towers

$$y = [f(x)]^{g(x)} \implies \frac{1}{y} \frac{dy}{dx} = g'(x)\ln f(x) + g(x)\frac{f'(x)}{f(x)}$$

Rule: Take natural logarithms on both sides before differentiating functions of the form $(\sin x)^{\cos x}$.

Application: Standard pattern in DSSSB PGT Mathematics for functions raised to trigonometric powers.
03

Inverse Trigonometric Simplification

$$\frac{d}{dx}\left[\sin^{-1}\left(\frac{2x}{1+x^2}\right)\right] = \frac{2}{1+x^2} \quad (\text{for } |x| \le 1)$$

Rule: Substitute $x = \tan \theta$ to simplify inverse trigonometric expressions algebraically before differentiating.

Application: Extremely frequent in UP TGT/PGT to bypass lengthy chain rule calculations.
04

nth Derivative of Exponential-Trigonometric Products

$$y = e^{ax} \sin(bx + c) \implies y_n = (a^2 + b^2)^{n/2} e^{ax} \sin(bx + c + n \tan^{-1}\frac{b}{a})$$

Rule: Use the standard $n$-th derivative formula directly instead of repeated differentiation.

05

Parametric Trigonometric Differentiation

$$x = a(\theta - \sin \theta), \, y = a(1 - \cos \theta) \implies \frac{dy}{dx} = \tan\left(\frac{\theta}{2}\right)$$

Rule: Divide $\frac{dy}{d\theta}$ by $\frac{dx}{d\theta}$ and use half-angle identities to simplify instantly.

06

Rapid Elimination via Domain & Periodicity

$$f(x+T) = f(x) \implies f'(x+T) = f'(x)$$

Rule: If a function is periodic, its derivative shares the same or related periodicity. Eliminate options with incorrect signs or frequencies.

Common errors and counterexamples

These examples show how a conclusion can fail when a condition is missing:

⚠️ COMMON ERROR #01

Sign Error in Cofunction Derivatives

Common Pitfall: Forgetting the negative sign when differentiating cosine, cotangent, and cosecant functions (e.g., writing $\frac{d}{dx}(\cos x) = \sin x$).

Counterexample: \frac{d}{dx}(\cos x) = -\sin x \neq \sin x

Why it fails: Violates the fundamental limit definition where $\cos(x+h) - \cos x = -2\sin(x+h/2)\sin(h/2)$.

🛡️ Exam Defense: Remember: All derivatives of co-functions (starting with 'co-') carry a negative sign.
⚠️ COMMON ERROR #02

Ignoring Domain Restrictions in Inverse Trig Derivatives

Common Pitfall: Applying derivative formulas for inverse trigonometric functions outside their valid principal domains.

Counterexample: \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}} \text{ only for } |x| < 1

Why it fails: At $x = \pm 1$, the denominator becomes zero, resulting in a vertical tangent where the derivative is undefined.

🛡️ Exam Defense: Always check boundary values of the argument before asserting differentiability.

Comparison

Concept / FunctionDomain / ConditionsDerivative FormulaGeometric MeaningExam Identification Clue
Sine Functionx \in \mathbb{R}\frac{d}{dx}(\sin x) = \cos xSlope of sine wave oscillationDirect formula or simple chain rule
Tangent Functionx \neq \frac{\pi}{2} + k\pi\frac{d}{dx}(\tan x) = \sec^2 xSquare of secant representing growth rateQuotient rule or trigonometric identity simplification
Inverse Sinex \in (-1, 1)\frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1-x^2}}Slope of inverse sinusoidal arcSubstitution with $\sin \theta$ or $\cos \theta$
Secant Functionx \neq \frac{\pi}{2} + k\pi\frac{d}{dx}(\sec x) = \sec x \tan xProduct of secant and tangent slopesReciprocal rule application

Solved examples

Each problem includes a full solution and a shorter method:

CANONICAL PROBLEM 01 Dual-Method Breakdown

If $y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \dots \to \infty}}}$, then $\frac{dy}{dx}$ is equal to:

(A) \frac{\cos x}{2y - 1}(B) \frac{\sin x}{2y - 1}(C) \frac{\cos x}{2y + 1}(D) \frac{-\cos x}{2y - 1}
✓ Correct Answer: (A)
📚 Method A: Formal Textbook Derivation
Step 1: Write $y = \sqrt{\sin x + y}$. Step 2: Square both sides to get $y^2 = \sin x + y$. Step 3: Differentiate implicitly with respect to $x$: $2y \frac{dy}{dx} = \cos x + \frac{dy}{dx}$. Step 4: Rearrange to get $\frac{dy}{dx}(2y - 1) = \cos x \implies \frac{dy}{dx} = \frac{\cos x}{2y - 1}$.
⚡ Method B: 25-Second Exam Shortcut
Use infinite series derivative formula: for $y = \sqrt{f(x) + y}$, $\frac{dy}{dx} = \frac{f'(x)}{2y - 1}$.
💡 Tip: Recognize infinite nesting structures immediately to write the answer.
CANONICAL PROBLEM 02 Dual-Method Breakdown

If $y = \tan^{-1}\left(\frac{\cos x - \sin x}{\cos x + \sin x}\right)$, then $\frac{dy}{dx}$ is:

(A) 1(B) -1(C) \frac{1}{1+x^2}(D) \frac{-1}{1+x^2}
✓ Correct Answer: (B)
📚 Method A: Formal Textbook Derivation
Step 1: Divide numerator and denominator by $\cos x$ to get $\tan^{-1}\left(\frac{1 - \tan x}{1 + \tan x}\right)$. Step 2: Express as $\tan^{-1}\left(\tan(\frac{\pi}{4} - x)\right)$. Step 3: Simplify to $y = \frac{\pi}{4} - x$. Step 4: Differentiate to obtain $\frac{dy}{dx} = -1$.
⚡ Method B: 25-Second Exam Shortcut
Standard trigonometric reduction: $\frac{\cos x - \sin x}{\cos x + \sin x} = \tan(\frac{\pi}{4} - x)$.
💡 Tip: Look for linear combinations of sine and cosine inside inverse tangent; they always reduce to $\frac{\pi}{4} \pm x$.

Practice questions

Choose an answer to check it and see the solution:

Q1
MCQ 1: If $f(x) = \log_e(\tan x)$, then $f'\left(\frac{\pi}{4}\right)$ is equal to:
A 0
B 1
C 2
D $\sqrt{2}$
Derivation & Analysis:

Given $f(x) = \ln(\tan x)$. By chain rule, $f'(x) = \frac{1}{\tan x} \cdot \sec^2 x = \frac{\cos x}{\sin x} \cdot \frac{1}{\cos^2 x} = \frac{1}{\sin x \cos x} = \frac{2}{\sin 2x}$. At $x = \frac{\pi}{4}$, $f'(\frac{\pi}{4}) = \frac{2}{\sin(\pi/2)} = 2$.

Q2
MCQ 2: The derivative of $\sin(x^2)$ with respect to $x^2$ is:
A $\cos(x^2)$
B $2x \cos(x^2)$
C $-\cos(x^2)$
D $\sin(x^2)$
Derivation & Analysis:

Let $u = x^2$. We need to find $\frac{d}{du}(\sin u) = \cos u = \cos(x^2)$. Alternatively, $\frac{d/dx(\sin(x^2))}{d/dx(x^2)} = \frac{2x \cos(x^2)}{2x} = \cos(x^2)$.

Q3
MCQ 3: If $y = \sec(\tan^{-1} x)$, then $\frac{dy}{dx}$ at $x = 1$ is:
A $\frac{1}{\sqrt{2}}$
B $\sqrt{2}$
C 1
D 2
Derivation & Analysis:

Let $\theta = \tan^{-1} x$, so $\tan \theta = x$ and $\sec \theta = \sqrt{1 + \tan^2 \theta} = \sqrt{1 + x^2}$. Thus $y = \sqrt{1 + x^2}$. Differentiating with respect to $x$, $\frac{dy}{dx} = \frac{x}{\sqrt{1+x^2}}$. At $x = 1$, $\frac{1}{\sqrt{2}}$? Wait, $\sec(\theta) = \sqrt{1+x^2}$, derivative is $\frac{x}{\sqrt{1+x^2}}$. At $x=1$, value is $\frac{1}{\sqrt{2}}$? Let us re-verify: $\frac{d}{dx}(\sec(\tan^{-1} x)) = \sec(\tan^{-1}x)\tan(\tan^{-1}x) \cdot \frac{1}{1+x^2} = \frac{\sqrt{1+x^2} \cdot x}{1+x^2} = \frac{x}{\sqrt{1+x^2}}$. At $x=1$, it is $\frac{1}{\sqrt{2}}$. Correct option is (A).

Q4
MCQ 4: If $y = \sin^{-1} x + \cos^{-1} x$, then $\frac{dy}{dx}$ is equal to:
A $\frac{2}{\sqrt{1-x^2}}$
B 0
C 1
D -1
Derivation & Analysis:

We know that $\sin^{-1} x + \cos^{-1} x = \frac{\pi}{2}$ for all $x \in [-1, 1]$. Differentiating a constant gives $\frac{dy}{dx} = 0$.

Q5
MCQ 5: If $y = e^{\tan x}$, then $\frac{dy}{dx}$ is:
A $e^{\tan x}$
B $e^{\tan x} \sec^2 x$
C $\sec^2 x$
D $e^{\tan x} \tan x$
Derivation & Analysis:

By the chain rule, $\frac{dy}{dx} = e^{\tan x} \cdot \frac{d}{dx}(\tan x) = e^{\tan x} \sec^2 x$.

Q6
MCQ 6: The 10th derivative of $\cos x$ with respect to $x$ is:
A $\cos x$
B $-\cos x$
C $\sin x$
D $-\sin x$
Derivation & Analysis:

The derivatives of $\cos x$ cycle every 4 steps: $-\sin x, -\cos x, \sin x, \cos x$. For $n=10$, $10 = 4(2) + 2$, which corresponds to the second derivative, $-\cos x$.

Q7
MCQ 7: If $x = \theta + \sin \theta$ and $y = 1 - \cos \theta$, then $\frac{dy}{dx}$ at $\theta = \frac{\pi}{2}$ is:
A 0
B 1
C -1
D $\frac{1}{2}$
Derivation & Analysis:

$\frac{dy}{d\theta} = \sin \theta$, $\frac{dx}{d\theta} = 1 + \cos \theta$. Thus $\frac{dy}{dx} = \frac{\sin \theta}{1 + \cos \theta} = \tan(\theta/2)$. At $\theta = \frac{\pi}{2}$, $\tan(\pi/4) = 1$.

Q8
MCQ 8: Assertion (A): $\frac{d}{dx}(\tan^{-1} x + \cot^{-1} x) = 0$. Reason (R): $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ for all $x \in \mathbb{R}$.
A Both (A) and (R) are true and (R) is the correct explanation of (A)
B Both (A) and (R) are true but (R) is not the correct explanation of (A)
C (A) is true but (R) is false
D (A) is false but (R) is true
Derivation & Analysis:

The identity $\tan^{-1} x + \cot^{-1} x = \frac{\pi}{2}$ holds for all real $x$, and its derivative is identically zero.

Q9
MCQ 9: If $y = \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}$, then $\frac{dy}{dx}$ is:
A $\sec^2 x$
B $-\sec^2 x$
C $\csc^2 x$
D $2\sec^2 x$
Derivation & Analysis:

Using trigonometric identities, $\frac{1-\cos 2x}{1+\cos 2x} = \frac{2\sin^2 x}{2\cos^2 x} = \tan^2 x$. Thus $y = \sqrt{\tan^2 x} = |\tan x|$. For principal domain, $y = \tan x$, and $\frac{dy}{dx} = \sec^2 x$.

Q10
MCQ 10: If $y = \sin(\ln x)$, then $x^2 y_2 + x y_1$ is equal to:
A y
B -y
C 0
D 1
Derivation & Analysis:

Let $y = \sin(\ln x)$. $y_1 = \cos(\ln x) \cdot \frac{1}{x} \implies x y_1 = \cos(\ln x)$. Differentiating again: $x y_2 + y_1 = -\sin(\ln x) \cdot \frac{1}{x} \implies x^2 y_2 + x y_1 = -\sin(\ln x) = -y$.

Common questions

The negative sign arises in co-functions (like $\cos^{-1} x$, $\cot^{-1} x$, and $\csc^{-1} x$) due to the strictly decreasing nature of these inverse functions over their principal domains, reflecting a negative instantaneous rate of change.

Forgetting to apply the Chain Rule to the interior argument (e.g., writing $\frac{d}{dx}(\sin(2x)) = \cos(2x)$ instead of $2\cos(2x)$).

Mastering standard trigonometric identities to simplify expressions before differentiation (e.g., half-angle substitutions) bypasses cumbersome quotient and product rules entirely.

← Previous Topic
Tracing of simple curves
Next Topic →
Degree of differential equations