MathsISC Class 12 · Continuity, Differentiability and Differentiation

Implicit Differentiation and Derivatives of Inverse Trigonometric Functions

Q: When should I use implicit differentiation?

Use it whenever x and y are mixed in a relation and it is awkward or impossible to make y the subject, such as x² + 3xy + y² = 11 or sin(x + y) = xy. It also gives the derivatives of inverse trigonometric functions.

Q: Why simplify inverse trigonometric functions before differentiating?

A substitution such as x = tan θ often reduces an expression like sin⁻¹(2x/(1 + x²)) to 2tan⁻¹x, whose derivative is immediate. Differentiating the original form directly is much longer and error-prone.

Q: What is logarithmic differentiation?

It is taking the logarithm of both sides before differentiating. It is needed when the variable appears in both the base and the exponent, as in xˣ or x^y = y^(2x), and it also simplifies long products and quotients.

Q: Do I need to remember the derivatives of sec⁻¹x and cosec⁻¹x?

Yes, they are part of the derivatives of inverse trigonometric functions in the syllabus, though sin⁻¹, cos⁻¹ and tan⁻¹ appear far more often in questions.

Q: Is differentiation by substitution in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus lists derivatives of inverse trigonometric functions with differentiation by means of substitution, derivatives of implicit functions, logarithmic differentiation, and differentiation of one function with respect to another.

Updated 15 October 2026 · Fact-checked

In implicit differentiation you differentiate both sides of a relation in x and y with respect to x, multiplying each derivative of a y-term by dy/dx, then solve for dy/dx. For inverse trigonometric functions, first simplify with a substitution such as x = tan θ, then use d/dx(tan⁻¹x) = 1/(1 + x²) and similar results.

Read cos y from the triangle (adjacent over hypotenuse = √(1 − x²)), so dy/dx = 1/cos y gives 1/√(1 − x²).

Illustration drawn for QuestionBuddy.

Understand Implicit and Inverse Trigonometric Differentiation

Implicit functions. A relation such as $x^{2} + 3 x y + y^{2} = 11$ defines $y$ as a function of $x$ near each point, even though you cannot easily write $y = \dots$ . To find $\frac{d y}{d x}$ , differentiate every term with respect to $x$ , treating $y$ as a function of $x$ . By the chain rule, $\frac{d}{d x} (y^{2}) = 2 y \frac{d y}{d x}$ , $\frac{d}{d x} (sin y) = cos y \frac{d y}{d x}$ , and a product like $x y$ needs the product rule: $\frac{d}{d x} (x y) = y + x \frac{d y}{d x}$ . Then collect the $\frac{d y}{d x}$ terms on one side and factor.

Derivatives of inverse trigonometric functions. These follow from implicit differentiation. If $y = sin^{- 1} x$ , then $sin y = x$ , so $cos y \frac{d y}{d x} = 1$ and $\frac{d y}{d x} = \frac{1}{c o s y} = \frac{1}{1 - x ^{2}}$ (positive, because $y \in [- \frac{π}{2}, \frac{π}{2}]$ ). In the same way:

$\frac{d}{d x} sin^{- 1} x = \frac{1}{1 - x ^{2}}$ , $\frac{d}{d x} cos^{- 1} x = - \frac{1}{1 - x ^{2}}$
$\frac{d}{d x} tan^{- 1} x = \frac{1}{1 + x ^{2}}$ , $\frac{d}{d x} cot^{- 1} x = - \frac{1}{1 + x ^{2}}$
$\frac{d}{d x} sec^{- 1} x = \frac{1}{∣ x ∣ x ^{2} - 1}$ , $\frac{d}{d x} cosec^{- 1} x = - \frac{1}{∣ x ∣ x ^{2} - 1}$

With the chain rule, $\frac{d}{d x} tan^{- 1} (u) = \frac{1}{1 + u ^{2}} \frac{d u}{d x}$ .

Simplify before you differentiate. Many ISC questions put a complicated algebraic argument inside an inverse function. Differentiating it directly is long and error-prone; the ISC 2025 examiners noted that candidates did not simplify using inverse trigonometric properties before differentiating. Recognise the patterns:

$\frac{2 x}{1 + x ^{2}}$ , $\frac{1 - x ^{2}}{1 + x ^{2}}$ , $\frac{2 x}{1 - x ^{2}}$ : put $x = tan θ$ ; they become $sin 2 θ$ , $cos 2 θ$ , $tan 2 θ$ .
$1 - x^{2}$ or $3 x - 4 x^{3}$ : put $x = sin θ$ (or $cos θ$ ).
$\frac{1 + x ^{2} - 1}{x}$ : put $x = tan θ$ ; it becomes $tan \frac{θ}{2}$ .

After simplifying, the function is often a multiple of $θ$ , that is, of $tan^{- 1} x$ or $sin^{- 1} x$ , whose derivative is immediate. Check the range of $θ$ so the simplification is valid; for example $sin^{- 1} \frac{2 x}{1 + x ^{2}} = 2 tan^{- 1} x$ only for $∣ x ∣ \leq 1$ .

Logarithmic differentiation. When the variable appears in an exponent ( $x^{x}$ , $x^{y} = y^{x}$ , $x^{y} = e^{x - y}$ ), take $lo g$ of both sides first. This turns powers into products, which are easy to differentiate. For $x^{y} = y^{x}$ : $y lo g x = x lo g y$ , and implicit differentiation finishes it.

Differentiating one function with respect to another. To differentiate $u (x)$ with respect to $v (x)$ , compute $\frac{d u}{d v} = \frac{d u / d x}{d v / d x}$ . The syllabus example is differentiating $sin x^{3}$ with respect to $x^{3}$ , which gives $cos x^{3}$ directly.

Proof questions with a condition. A relation like $1 - x^{2} + 1 - y^{2} = a (x - y)$ is best handled by substitution ( $x = sin A$ , $y = sin B$ ), which reduces it to $sin^{- 1} x - sin^{- 1} y = constant$ ; then the derivative is one line.

Implicit and Inverse Trigonometric Differentiation: formulas to remember

Implicit chain rule: $\frac{d}{d x} [g (y)] = g^{'} (y) \frac{d y}{d x}$; Every y-term picks up a dy/dx.
sin⁻¹ and cos⁻¹: $\frac{d}{d x} sin^{- 1} x = \frac{1}{1 - x ^{2}}, \frac{d}{d x} cos^{- 1} x = \frac{- 1}{1 - x ^{2}}$; |x| < 1.
tan⁻¹ and cot⁻¹: $\frac{d}{d x} tan^{- 1} x = \frac{1}{1 + x ^{2}}, \frac{d}{d x} cot^{- 1} x = \frac{- 1}{1 + x ^{2}}$; All real x.
sec⁻¹ and cosec⁻¹: $\frac{d}{d x} sec^{- 1} x = \frac{1}{∣ x ∣ x ^{2} - 1}, \frac{d}{d x} cosec^{- 1} x = \frac{- 1}{∣ x ∣ x ^{2} - 1}$; |x| > 1.
tan substitution: $x = tan θ : \frac{2 x}{1 + x ^{2}} = sin 2 θ, \frac{1 - x ^{2}}{1 + x ^{2}} = cos 2 θ, \frac{2 x}{1 - x ^{2}} = tan 2 θ$; Turns the inverse function into 2θ = 2tan⁻¹x (within the right range).
Logarithmic differentiation: $y = u^{v} \Rightarrow lo g y = v lo g u$; Use whenever the variable is in the exponent.
One function with respect to another: $\frac{d u}{d v} = \frac{d u / d x}{d v / d x}$; dv/dx ≠ 0.

How to solve Implicit and Inverse Trigonometric Differentiation questions

Decide first which kind of question it is: an implicit relation, an inverse trigonometric expression, or a variable exponent. Then follow these steps.

1If the variable is in an exponent, take logarithms of both sides first.
2If an inverse trigonometric function has an algebraic argument, look for a substitution ( $x = tan θ$ , $sin θ$ or $cos θ$ ) that simplifies it, and note the range of $θ$ .
3Rewrite the function in its simplest form (for example $y = 2 tan^{- 1} x$ ) before differentiating.
4For an implicit relation, differentiate each term with respect to $x$ , writing $\frac{d y}{d x}$ after every derivative of a $y$ -term; use the product rule on terms like $x y$ .
5Collect all $\frac{d y}{d x}$ terms on one side, factor out $\frac{d y}{d x}$ , and divide.
6If the answer must be in a given form, substitute back from the original relation (for example replace $y$ using $y = \frac{x}{1 + l o g x}$ ).
7If a numerical slope is asked, substitute the point only at the end, after checking it satisfies the relation.

Implicit and Inverse Trigonometric Differentiation in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Implicit differentiation for the slope of a curve

Suppose a circular running track has radius 20 m. With its centre at the origin, the track is $x^{2} + y^{2} = 400$ . A runner is at the point $(12, 16)$ .

Find the slope of the direction in which the runner is moving at that point.

Try it first, then show the solutionSolution

Differentiate implicitly: $2 x + 2 y \frac{d y}{d x} = 0$ , so $\frac{d y}{d x} = - \frac{x}{y}$ .
At $(12, 16)$ : $\frac{d y}{d x} = - \frac{12}{16} = - \frac{3}{4}$ .
This is the slope of the tangent, which is the direction of motion; it is perpendicular to the radius, whose slope is $\frac{16}{12} = \frac{4}{3}$ .

Answer: The runner moves along a line of slope $- \frac{3}{4}$ .

Memory hook: On a round track, the running direction is the tangent: dy/dx = −x/y.

Derivative of tan⁻¹

Imagine a camera on the ground that tracks a drone flying horizontally at a height of 50 m. When the drone is $x$ m (horizontally) past the point directly above the camera, the camera tilts by $θ = tan^{- 1} \frac{x}{50}$ from the vertical.

Find $\frac{d θ}{d x}$ , and its value when $x = 50$ m.

Try it first, then show the solutionSolution

$\frac{d θ}{d x} = \frac{1}{1 + ( x /50 ) ^{2}} \cdot \frac{1}{50} = \frac{50}{2500 + x ^{2}}$ .
At $x = 50$ : $\frac{50}{5000} = \frac{1}{100}$ radian per metre.
So when the drone is 50 m past the camera, each extra metre it flies tilts the camera by about 0.01 rad (about $0.5 7^{\circ}$ ).

Answer: $\frac{d θ}{d x} = \frac{50}{2500 + x ^{2}}$ ; at $x = 50$ m it is $\frac{1}{100}$ rad per metre.

Memory hook: Tracking angles shrink their rate as the drone goes further: 1/(1 + u²) in action.

Tricks to remember Implicit and Inverse Trigonometric Differentiation

Short memory aids. Say each one aloud once, then use it on a question the same day.

Every y gets a tag
When differentiating with respect to x, every time you differentiate something with y in it, attach a dy/dx tag.
Use it when: Avoiding the commonest implicit-differentiation error.
Co-functions take a minus
The derivatives of cos⁻¹, cot⁻¹ and cosec⁻¹ are the negatives of those of sin⁻¹, tan⁻¹ and sec⁻¹.
Use it when: Recalling six derivatives from three.
1 + x² means tan
See 1 + x² in an inverse function? Put x = tan θ. See √(1 − x²)? Put x = sin θ.
Use it when: Choosing the substitution in seconds.
Exponent with x? Take logs
xˣ, x^y and similar: log first, then differentiate.
Use it when: Logarithmic differentiation questions.

Implicit and Inverse Trigonometric Differentiation: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Slope from an implicit relation · 2 marks · Short answerBoard level

If $x^{2} + 3 x y + y^{2} = 11$ , find $\frac{d y}{d x}$ and its value at $(1, 2)$ .

Try it first, then show the solutionSolution

Differentiate: $2 x + 3 (y + x \frac{d y}{d x}) + 2 y \frac{d y}{d x} = 0$ . (1 mark)
$\frac{d y}{d x} (3 x + 2 y) = - (2 x + 3 y) \Rightarrow \frac{d y}{d x} = - \frac{2 x + 3 y}{3 x + 2 y}$ . At $(1, 2)$ : $- \frac{8}{7}$ . (1 mark)

Answer: $\frac{d y}{d x} = - \frac{2 x + 3 y}{3 x + 2 y}$ ; at $(1, 2)$ it is $- \frac{8}{7}$ .

Example 2: Chain rule with tan⁻¹ · 1 mark · MCQBoard level

The derivative of $tan^{- 1} (x^{2})$ with respect to $x$ is:
(a) $\frac{1}{1 + x ^{4}}$
(b) $\frac{2 x}{1 + x ^{4}}$
(c) $\frac{2 x}{1 + x ^{2}}$
(d) $\frac{1}{1 + x ^{2}}$

Try it first, then show the solutionSolution

$\frac{1}{1 + ( x ^{2} ) ^{2}} \cdot 2 x$ .

Answer: (b) $\frac{2 x}{1 + x ^{4}}$ .

Example 3: Simplify, then differentiate · 3 marks · Short answerHarder board level

Differentiate $y = cos^{- 1} (4 x^{3} - 3 x)$ with respect to $x$ , for $\frac{1}{2} < x < 1$ .

Try it first, then show the solutionSolution

Put $x = cos θ$ . For $\frac{1}{2} < x < 1$ , $θ \in (0, \frac{π}{3})$ , and $4 x^{3} - 3 x = 4 cos^{3} θ - 3 cos θ = cos 3 θ$ with $3 θ \in (0, π)$ . (1 mark)
So $y = cos^{- 1} (cos 3 θ) = 3 θ = 3 cos^{- 1} x$ . (1 mark)
$\frac{d y}{d x} = - \frac{3}{1 - x ^{2}}$ . (1 mark)

Answer: $\frac{d y}{d x} = - \frac{3}{1 - x ^{2}}$ . (The same simplify-first idea, with $\frac{2 \cdot 6 ^{x}}{1 + 3 6 ^{x}}$ inside $sin^{- 1}$ , was tested in ISC 2025.)

Example 4: A half-angle substitution · 3 marks · Short answerHarder board level

Differentiate $y = tan^{- 1} (\frac{1 + x ^{2} - 1}{x})$ , $x \neq = 0$ .

Try it first, then show the solutionSolution

Put $x = tan θ$ , $θ \in (- \frac{π}{2}, \frac{π}{2})$ , $θ \neq = 0$ . Then $1 + x^{2} = sec θ$ . (1 mark)
$\frac{s e c θ - 1}{t a n θ} = \frac{1 - c o s θ}{s i n θ} = \frac{2 s i n ^{2} ( θ /2 )}{2 s i n ( θ /2 ) c o s ( θ /2 )} = tan \frac{θ}{2}$ . (1 mark)
So $y = \frac{θ}{2} = \frac{1}{2} tan^{- 1} x$ and $\frac{d y}{d x} = \frac{1}{2 ( 1 + x ^{2} )}$ . (1 mark)

Answer: $\frac{d y}{d x} = \frac{1}{2 ( 1 + x ^{2} )}$ .

Example 5: Logarithmic and implicit together · 3 marks · Short answerHarder board level

If $x^{y} = y^{2 x}$ , find $\frac{d y}{d x}$ .

Try it first, then show the solutionSolution

Take logs: $y lo g x = 2 x lo g y$ . (½ mark)
Differentiate: $\frac{d y}{d x} lo g x + \frac{y}{x} = 2 lo g y + \frac{2 x}{y} \frac{d y}{d x}$ . (1 mark)
$\frac{d y}{d x} (lo g x - \frac{2 x}{y}) = 2 lo g y - \frac{y}{x}$ . (½ mark)
$\frac{d y}{d x} = \frac{2 l o g y - \frac{y}{x}}{l o g x - \frac{2 x}{y}} = \frac{y ( 2 x l o g y - y )}{x ( y l o g x - 2 x )}$ . (1 mark)

Answer: $\frac{d y}{d x} = \frac{y ( 2 x l o g y - y )}{x ( y l o g x - 2 x )}$ . At the point $(2, 16)$ , which satisfies $2^{16} = 1 6^{4}$ , this is about $- 5.54$ .

Example 6: A relation solved by substitution · 5 marks · Long answerHOTS / competency

If $1 - x^{2} + 1 - y^{2} = a (x - y)$ , prove that $\frac{d y}{d x} = \frac{1 - y ^{2}}{1 - x ^{2}}$ .

Try it first, then show the solutionSolution

Put $x = sin A$ , $y = sin B$ (with $A, B$ in $[- \frac{π}{2}, \frac{π}{2}]$ ). Then $cos A + cos B = a (sin A - sin B)$ . (1 mark)
$2 cos \frac{A + B}{2} cos \frac{A - B}{2} = 2 a cos \frac{A + B}{2} sin \frac{A - B}{2}$ . (1 mark)
Cancelling $2 cos \frac{A + B}{2}$ (non-zero except when $x = y = \pm 1$ , which we exclude): $cot \frac{A - B}{2} = a$ , so $A - B = 2 cot^{- 1} a$ , a constant. (1 mark)
So $sin^{- 1} x - sin^{- 1} y = 2 cot^{- 1} a$ . Differentiate: $\frac{1}{1 - x ^{2}} - \frac{1}{1 - y ^{2}} \frac{d y}{d x} = 0$ . (1 mark)
$\frac{d y}{d x} = \frac{1 - y ^{2}}{1 - x ^{2}} = \frac{1 - y ^{2}}{1 - x ^{2}}$ . (1 mark)

Answer: Proved.

Where students lose marks in Implicit and Inverse Trigonometric Differentiation

The slips examiners see most often, and what to write instead.

Mistake: Forgetting dy/dx on y-terms
Differentiating y² as 2y instead of 2y(dy/dx) treats y as a constant.
Wrong working
$\frac{d}{d x} (x^{2} + y^{2}) = 2 x + 2 y$ .
Right working
$\frac{d}{d x} (x^{2} + y^{2}) = 2 x + 2 y \frac{d y}{d x}$ .
Do this: Attach dy/dx every time you differentiate an expression in y.
Usually costs: The whole question.
Mistake: Differentiating before simplifying
The ISC 2025 examiners noted that candidates did not simplify the inverse trigonometric expression first and then misapplied the chain rule.
Do this: Look for a tan, sin or cos substitution and reduce the function to a multiple of tan⁻¹x or sin⁻¹x first.
Usually costs: 1 to 2 marks.
Mistake: Wrong log properties
In ISC 2025, candidates applied properties of logarithms incorrectly in x^y = e^(x − y).
Wrong working
$lo g (x^{y}) = (lo g x)^{y}$ .
Right working
$lo g (x^{y}) = y lo g x$ and $lo g (e^{x - y}) = x - y$ .
Do this: Write the log rule you are using in the margin.
Usually costs: The method mark and the answer.
Mistake: Missing the product rule on xy
xy depends on x in both factors.
Wrong working
$\frac{d}{d x} (x y) = \frac{d y}{d x}$ .
Right working
$\frac{d}{d x} (x y) = y + x \frac{d y}{d x}$ .
Do this: Treat any product of x and y as a product of two functions of x.
Usually costs: 1 mark.

Exam tips

When a proof asks for a specific form, keep the target in view and substitute from the original relation at the end.
Write the substitution and the range of θ in one line ("put $x = tan θ$ , $θ \in (- \frac{π}{4}, \frac{π}{4})$ "); it earns a mark and justifies the simplification.
For $x^{y}$ -type questions, the first line should always be "taking logarithms on both sides".
Leave $\frac{d y}{d x}$ in terms of $x$ and $y$ unless the question asks otherwise.
Check a numerical slope by substituting the point into the original relation first; if it does not satisfy it, re-read the question.

Practice questions on Implicit and Inverse Trigonometric Differentiation

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The derivative of $cos^{- 1} x$ for $∣ x ∣ < 1$ is:
(a) $\frac{1}{1 - x ^{2}}$
(b) $\frac{- 1}{1 - x ^{2}}$
(c) $\frac{1}{1 + x ^{2}}$
(d) $\frac{- 1}{1 + x ^{2}}$
Hint
Co-function: minus sign.
Answer
(b).
Q2 · 2 marks · Short answerBoard level
If $sin (x + y) = x y$ , find $\frac{d y}{d x}$ .
Hint
Chain rule on the left, product rule on the right.
Answer
$cos (x + y) (1 + \frac{d y}{d x}) = y + x \frac{d y}{d x}$ , so $\frac{d y}{d x} = \frac{y - c o s ( x + y )}{c o s ( x + y ) - x}$ .
Q3 · 3 marks · Short answerHarder board level
Differentiate $cos^{- 1} (\frac{1 - x ^{2}}{1 + x ^{2}})$ for $0 < x < 1$ .
Hint
Put $x = tan θ$ .
Answer
With $x = tan θ$ , $θ \in (0, \frac{π}{4})$ , the function is $cos^{- 1} (cos 2 θ) = 2 θ = 2 tan^{- 1} x$ , so the derivative is $\frac{2}{1 + x ^{2}}$ .
Q4 · 2 marks · Short answerHarder board level
Differentiate $sin^{2} x$ with respect to $e^{c o s x}$ .
Hint
Divide the two $x$ -derivatives.
Answer
$\frac{2 s i n x c o s x}{- s i n x e ^{c o s x}} = - 2 cos x e^{- c o s x}$ (for $sin x \neq = 0$ ).
Q5 · 2 marks · Short answerHarder board level
If $e^{y} (x + 1) = 1$ , show that $\frac{d y}{d x} = - e^{y}$ .
Hint
Take logs or differentiate implicitly.
Answer
$y = - lo g (x + 1)$ , so $\frac{d y}{d x} = - \frac{1}{x + 1} = - e^{y}$ , since $e^{y} = \frac{1}{x + 1}$ .
Q6 · 3 marks · Short answerHOTS / competency
If $y = \frac{s i n ^{- 1} x}{1 - x ^{2}}$ , show that $(1 - x^{2}) \frac{d y}{d x} - x y = 1$ .
Hint
Write $y 1 - x^{2} = sin^{- 1} x$ first.
Answer
Differentiate $y 1 - x^{2} = sin^{- 1} x$ : $\frac{d y}{d x} 1 - x^{2} - \frac{x y}{1 - x ^{2}} = \frac{1}{1 - x ^{2}}$ . Multiply by $1 - x^{2}$ : $(1 - x^{2}) \frac{d y}{d x} - x y = 1$ .
Q7 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a circular fountain basin of radius 5 m is drawn on a plan as $x^{2} + y^{2} = 25$ .
(i) Find $\frac{d y}{d x}$ by implicit differentiation. (1 mark)
(ii) Find the slope of the rim at $(3, 4)$ . (1 mark)
(iii) At which points of the rim is the tangent horizontal? (1 mark)
(iv) Check your answer to (ii) using the explicit form $y = 25 - x^{2}$ . (1 mark)
Hint
A horizontal tangent has slope 0.
Answer
(i) $- \frac{x}{y}$ . (ii) $- \frac{3}{4}$ . (iii) Where $x = 0$ : $(0, 5)$ and $(0, - 5)$ . (iv) $\frac{d y}{d x} = \frac{- x}{25 - x ^{2}} = - \frac{3}{4}$ at $x = 3$ .

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the derivative of $tan^{- 1} x$ .Show answer
$\frac{1}{1 + x ^{2}}$ .
2.What is $\frac{d}{d x} (y^{3})$ when $y$ is a function of $x$ ?Show answer
$3 y^{2} \frac{d y}{d x}$ .
3.Which substitution simplifies $sin^{- 1} \frac{2 x}{1 + x ^{2}}$ ?Show answer
$x = tan θ$ .
4.Find $\frac{d y}{d x}$ if $x y = 4$ .Show answer
$- \frac{y}{x}$ .
5.How do you differentiate $u$ with respect to $v$ ?Show answer
$\frac{d u}{d v} = \frac{d u / d x}{d v / d x}$ .
6.What is the derivative of $sin x^{3}$ with respect to $x^{3}$ ?Show answer
$cos x^{3}$ .

Implicit and Inverse Trigonometric Differentiation: frequently asked questions

When should I use implicit differentiation?

Use it whenever x and y are mixed in a relation and it is awkward or impossible to make y the subject, such as x² + 3xy + y² = 11 or sin(x + y) = xy. It also gives the derivatives of inverse trigonometric functions.

Why simplify inverse trigonometric functions before differentiating?

A substitution such as x = tan θ often reduces an expression like sin⁻¹(2x/(1 + x²)) to 2tan⁻¹x, whose derivative is immediate. Differentiating the original form directly is much longer and error-prone.

What is logarithmic differentiation?

It is taking the logarithm of both sides before differentiating. It is needed when the variable appears in both the base and the exponent, as in xˣ or x^y = y^(2x), and it also simplifies long products and quotients.

Do I need to remember the derivatives of sec⁻¹x and cosec⁻¹x?

Yes, they are part of the derivatives of inverse trigonometric functions in the syllabus, though sin⁻¹, cos⁻¹ and tan⁻¹ appear far more often in questions.

Is differentiation by substitution in the ISC 2027 syllabus?

Yes. The CISCE 2027 syllabus lists derivatives of inverse trigonometric functions with differentiation by means of substitution, derivatives of implicit functions, logarithmic differentiation, and differentiation of one function with respect to another.