Explicit vs Implicit Functions
Before learning implicit differentiation, it's important to understand the difference between explicit and implicit forms.
📗 Explicit Function
y is directly written in terms of x. You can clearly see what y equals.
y = sin(x) + eˣ
📘 Implicit Function
x and y are mixed together. y is NOT isolated on one side.
x³ + y³ = 3xy
A function defined by an equation of the form F(x, y) = 0, where y is not explicitly expressed as a function of x alone, is called an implicit function.
When y cannot be separated easily, we differentiate both sides with respect to x — treating y as a function of x — and then solve for dy/dx.
The Key Rule — Differentiating y Terms
The core idea: whenever you differentiate a term containing y with respect to x, you must multiply by dy/dx (chain rule, since y is a function of x).
💡 Think of it this way: d/dx[f(y)] = f'(y) · dy/dx — it's just the chain rule where the inner function is y(x).
Step-by-Step Method
Differentiate both sides w.r.t. x
Apply d/dx to every term on both left and right sides of the equation.
Use chain rule on y-terms
Every y-term gets multiplied by dy/dx. Use product/quotient rules as needed.
Collect all dy/dx terms on one side
Move all terms containing dy/dx to the left, everything else to the right.
Factor out dy/dx
Write: dy/dx × (combined factor) = (rest of the expression).
Solve for dy/dx
Divide both sides by the factor to get dy/dx = ...
Solved Examples
Differentiate both sides w.r.t. x:
d/dx[x²] + d/dx[y²] = d/dx[25]
2x + 2y·(dy/dx) = 0
2y·(dy/dx) = −2x
dy/dx = −x/y
Differentiate both sides:
3x² + 3y²·(dy/dx) = 3[y + x·(dy/dx)]
3x² + 3y²·(dy/dx) = 3y + 3x·(dy/dx)
3y²·(dy/dx) − 3x·(dy/dx) = 3y − 3x²
dy/dx (3y² − 3x) = 3y − 3x²
dy/dx = (y − x²) / (y² − x)
Differentiate LHS: cos(x+y) · (1 + dy/dx)
Differentiate RHS using Product Rule: 2y·cos x·(dy/dx) + y²·(−sin x)
cos(x+y) + cos(x+y)·dy/dx = 2y·cos x·dy/dx − y²·sin x
Collect dy/dx terms: cos(x+y)·dy/dx − 2y·cos x·dy/dx = −y²·sin x − cos(x+y)
dy/dx [cos(x+y) − 2y·cos x] = −y²·sin x − cos(x+y)
dy/dx = [−y² sin x − cos(x+y)] / [cos(x+y) − 2y cos x]
Differentiate both sides:
eˣ + eʸ·(dy/dx) = e^(x+y)·(1 + dy/dx)
eˣ + eʸ·(dy/dx) = e^(x+y) + e^(x+y)·(dy/dx)
dy/dx·[eʸ − e^(x+y)] = e^(x+y) − eˣ
dy/dx = [e^(x+y) − eˣ] / [eʸ − e^(x+y)] = −eˣ/eʸ (simplified)
Second Order Implicit Derivative
Sometimes we need d²y/dx² from an implicit equation. The method is to differentiate dy/dx once more with respect to x, again using the chain rule wherever y appears.
We already know: dy/dx = −x/y
d²y/dx² = d/dx[−x/y] using Quotient Rule:
= [−y − (−x)·(dy/dx)] / y²
= [−y + x·(−x/y)] / y²
= [−y² − x²] / y³ = −(x² + y²)/y³
Since x²+y² = r²: d²y/dx² = −r²/y³
⚠️ When finding d²y/dx², always substitute the expression for dy/dx back into the result. Don't leave d²y/dx² in terms of dy/dx.
Chapter 5 — Complete Formula Sheet
f continuous at a ⟺ lim f(x) = f(a) = LHL = RHL
DIFFERENTIABILITY
f'(a) = lim [f(a+h)−f(a)]/h (h→0) | LHD = RHD
Diff ⟹ Continuous | Continuous ⟹ Diff (NOT always)
EXP & LOG
d/dx(eˣ) = eˣ | d/dx(aˣ) = aˣ ln a
d/dx(ln x) = 1/x | d/dx(logₐx) = 1/(x ln a)
LOG DIFFERENTIATION
y = f(x)^g(x) → ln y = g(x)·ln f(x) → (1/y)dy/dx = ...
INVERSE TRIG
d/dx(sin⁻¹x) = 1/√(1−x²) | d/dx(cos⁻¹x) = −1/√(1−x²)
d/dx(tan⁻¹x) = 1/(1+x²) | d/dx(cot⁻¹x) = −1/(1+x²)
d/dx(sec⁻¹x) = 1/|x|√(x²−1) | d/dx(cosec⁻¹x) = −1/|x|√(x²−1)
IMPLICIT DIFFERENTIATION
d/dx[F(x,y)=0] → differentiate every term → solve for dy/dx
d/dx[yⁿ] = n·yⁿ⁻¹·(dy/dx) ← always remember this!
🎯 Exam Tips: (1) Check LHL = RHL = f(a) for continuity. (2) Check LHD = RHD for differentiability. (3) For implicit: differentiate every term, collect dy/dx, factorise. (4) For log diff: take ln → simplify → differentiate → multiply back by y. (5) Inverse trig: memorise the 3 positive forms, negatives are their negatives.