On this page
Some relations between x and y cannot be rearranged into y = f(x), such as x³ + y³ = 3xy. Some functions, like x^x, fit no rule you know. Class 12 gives two tools for these: implicit and logarithmic differentiation. Both are really the chain rule used carefully. This post works through five examples, including a second derivative, and lists the traps that cost marks.
Where is this in the syllabus?
Both are in Chapter 5, Continuity and Differentiability, of the NCERT Class 12 mathematics textbook: section 5.3.2 for implicit functions and 5.5 for logarithmic differentiation. The CBSE 2026-27 curriculum lists "derivative of implicit functions" and "logarithmic differentiation". The JEE (Main) 2026 syllabus includes differentiation of implicit functions and "derivatives of order upto two", and the JEE (Advanced) 2026 brochure asks for "derivatives up to order two of implicit functions". Check the current year's documents.
How does implicit differentiation work?
Differentiate every term with respect to x, remembering that y depends on x. So:
d/dx (x²) = 2x
d/dx (y²) = 2y × dy/dx (chain rule)
d/dx (xy) = y + x dy/dx (product rule)
Then collect the dy/dx terms on one side and solve.
Worked example 1: a circle
x² + y² = 25. Differentiate: 2x + 2y dy/dx = 0, so dy/dx = −x/y.
At (3, 4) the slope is −3/4. Check it with geometry: the radius to (3, 4) has slope 4/3, and a tangent is perpendicular to the radius, so its slope is −3/4.
Worked example 2: the folium x³ + y³ = 3xy
Differentiate: 3x² + 3y² dy/dx = 3y + 3x dy/dx.
Divide by 3 and collect: dy/dx (y² − x) = y − x², so dy/dx = (y − x²) / (y² − x).
The point (3/2, 3/2) lies on the curve, since 27/8 + 27/8 = 27/4 = 3 × 9/4. There, dy/dx = (3/2 − 9/4) / (9/4 − 3/2) = (−3/4) / (3/4) = −1.
Worked example 3: a second derivative
JEE asks for second derivatives of implicit functions. Go back to the circle, where dy/dx = −x/y. Differentiate again with the quotient rule:
d²y/dx² = −(y − x dy/dx) / y²
Substitute dy/dx = −x/y: d²y/dx² = −(y + x²/y) / y² = −(x² + y²) / y³ = −25/y³, using the original equation.
At (3, 4): d²y/dx² = −25/64. Using the original equation to simplify is the step most students miss.
How does logarithmic differentiation work?
NCERT's section 5.5 handles functions of the form y = [u(x)]^v(x). Take natural logs, log y = v(x) log u(x), then differentiate both sides, using the chain rule on the left:
(1/y) dy/dx = v(x) u′(x)/u(x) + v′(x) log u(x)
NCERT adds the condition that matters: f(x) and u(x) "must always be positive as otherwise their logarithms are not defined".
Worked example 4: y = x^x, for x > 0
log y = x log x. Differentiate: (1/y) dy/dx = log x + 1. So dy/dx = x^x (1 + log x).
Check at x = 1: y = 1 and dy/dx = 1 × (1 + 0) = 1.
Two wrong answers appear on almost every paper: x × x^(x − 1) (the power rule, which needs a constant exponent) and x^x log x (the exponential rule, which needs a constant base). The correct answer is their sum, which is a good way to remember it.
Worked example 5: a long product
y = (x + 1)² (x − 2)³ / (x + 3)⁴, for x > 2.
log y = 2 log(x + 1) + 3 log(x − 2) − 4 log(x + 3), so
dy/dx = y [2/(x + 1) + 3/(x − 2) − 4/(x + 3)].
At x = 3: y = 16 × 1 / 1296 = 1/81, and the bracket is 1/2 + 3 − 2/3 = 17/6. So dy/dx = (1/81) × (17/6) = 17/486. Doing this with the product and quotient rules takes a page; the log turns it into three lines.
Why is the chain rule the real difficulty?
In interviews with 25 students who had just finished a single-variable calculus course, Borji and Martínez-Planell (2020) found that understanding the chain rule and the idea of an implicit function played a key role in whether students made sense of implicit differentiation. If you treat "differentiate y²" as a rule to memorise, rather than as the chain rule, the dy/dx gets lost. For where derivatives start, see derivatives from first principles.
Which mistakes cost marks?
Writing d/dx (y²) = 2y without dy/dx.
Forgetting the product rule on terms like xy or x sin y.
Taking the log of a sum. log(u + v) is not log u + log v. For y = x^x + (sin x)^x, differentiate each part separately and add.
Using the power rule or the exponential rule on x^x.
Not simplifying the second derivative with the original equation.
If one of these slips keeps coming back, the Fermi Problem Library has a step-by-step solver that explains the reasoning rather than printing an answer, so you can see exactly where a dy/dx or a product rule was dropped.
A practice routine
Do NCERT Exercise 5.3 (implicit) and 5.5 (logarithmic), circling every dy/dx.
Find d²y/dx² for x² + y² = r² and xy = c², simplifying with the original equation each time.
Differentiate x^x, x^(sin x) and (sin x)^(cos x), and check each at a convenient point.
Use your answers to find tangent lines, as in our post on tangents and normals.
Today, find dy/dx for xy + y² = 4 and check that it gives −1/5 at the point (3, 1).
