Sometimes y y y trapped inside an equation and refuses to come out cleanly. Try solving x 2 + y 2 = 25 x^2 + y^2 = 25 x 2 + y 2 = 25 y y y y = ± 25 − x 2 y = \pm\sqrt{25 - x^2} y = ± 25 − x 2 sin ( x y ) = x + y \sin(xy) = x + y sin ( x y ) = x + y y y y at all in closed form.
The trick: we don't need to free y y y y y y x x x y y y d y d x \frac{dy}{dx} d x d y
Definition Explicit vs Implicit
An explicit function gives y y y y = f ( x ) y = f(x) y = f ( x )
An implicit equation relates x x x y y y F ( x , y ) = 0 F(x,y)=0 F ( x , y ) = 0 x 2 + y 2 = 25 x^2+y^2=25 x 2 + y 2 = 25
Implicit differentiation is differentiating both sides of an implicit equation with respect to x x x treating y y y x x x , then solving for d y d x \dfrac{dy}{dx} d x d y
The whole machine rests on one assumption + one rule .
Assumption. Near a point, the curve F ( x , y ) = 0 F(x,y)=0 F ( x , y ) = 0 y y y some differentiable function y ( x ) y(x) y ( x ) ∂ F / ∂ y ≠ 0 \partial F/\partial y \neq 0 ∂ F / ∂ y = 0
Rule. The chain rule. If u = g ( y ) u = g(y) u = g ( y ) y = y ( x ) y=y(x) y = y ( x ) d d x g ( y ) = g ′ ( y ) ⋅ d y d x . \frac{d}{dx}\,g(y) = g'(y)\cdot \frac{dy}{dx}. d x d g ( y ) = g ′ ( y ) ⋅ d x d y .
So whenever you differentiate a term containing y y y d y d x \dfrac{dy}{dx} d x d y only difference from ordinary differentiation.
d y d x \frac{dy}{dx} d x d y d d x ( x 3 ) = 3 x 2 \frac{d}{dx}(x^3) = 3x^2 d x d ( x 3 ) = 3 x 2 x x x x x x d d x ( y 3 ) = 3 y 2 ⋅ d y d x \frac{d}{dx}(y^3) = 3y^2 \cdot \frac{dy}{dx} d x d ( y 3 ) = 3 y 2 ⋅ d x d y y y y d y d x \frac{dy}{dx} d x d y x x x y ′ y' y ′ y y y
Differentiate F ( x , y ) = 0 F(x,y)=0 F ( x , y ) = 0 x x x F x + F y d y d x = 0 ⟹ d y d x = − F x F y F_x + F_y\,\frac{dy}{dx} = 0 \quad\Longrightarrow\quad \boxed{\frac{dy}{dx} = -\frac{F_x}{F_y}} F x + F y d x d y = 0 ⟹ d x d y = − F y F x F x = ∂ F / ∂ x F_x = \partial F/\partial x F x = ∂ F / ∂ x F y = ∂ F / ∂ y F_y=\partial F/\partial y F y = ∂ F / ∂ y term-by-term method below is what you actually use.
Worked example Example 1 — Circlex 2 + y 2 = 25 x^2 + y^2 = 25 x 2 + y 2 = 25 d d x ( x 2 ) + d d x ( y 2 ) = d d x ( 25 ) \frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25) d x d ( x 2 ) + d x d ( y 2 ) = d x d ( 25 ) 2 x + 2 y d y d x = 0 2x + 2y\frac{dy}{dx} = 0 2 x + 2 y d x d y = 0 Why this step? d d x ( y 2 ) = 2 y ⋅ y ′ \frac{d}{dx}(y^2)=2y\cdot y' d x d ( y 2 ) = 2 y ⋅ y ′ y y y d y d x = − 2 x 2 y = − x y . \dfrac{dy}{dx} = -\dfrac{2x}{2y} = -\dfrac{x}{y}. d x d y = − 2 y 2 x = − y x .
Check vs explicit: top half is y = 25 − x 2 y=\sqrt{25-x^2} y = 25 − x 2 y ′ = − x 25 − x 2 = − x y y'=\frac{-x}{\sqrt{25-x^2}} = -\frac{x}{y} y ′ = 25 − x 2 − x = − y x both branches at once.
Worked example Example 2 — Product termx 3 + y 3 = 6 x y x^3 + y^3 = 6xy x 3 + y 3 = 6 x y 3 x 2 + 3 y 2 d y d x = 6 ( y + x d y d x ) 3x^2 + 3y^2\frac{dy}{dx} = 6\left(y + x\frac{dy}{dx}\right) 3 x 2 + 3 y 2 d x d y = 6 ( y + x d x d y ) Why this step? RHS 6 x y 6xy 6 x y d d x ( 6 x y ) = 6 ( y ⋅ 1 + x ⋅ y ′ ) \frac{d}{dx}(6xy)=6(y\cdot 1 + x\cdot y') d x d ( 6 x y ) = 6 ( y ⋅ 1 + x ⋅ y ′ ) product rule — and the x x x y ′ y' y ′ 3 y 2 y ′ − 6 x y ′ = 6 y − 3 x 2 ⇒ y ′ ( 3 y 2 − 6 x ) = 6 y − 3 x 2 3y^2 y' - 6x\,y' = 6y - 3x^2 \;\Rightarrow\; y'(3y^2-6x)=6y-3x^2 3 y 2 y ′ − 6 x y ′ = 6 y − 3 x 2 ⇒ y ′ ( 3 y 2 − 6 x ) = 6 y − 3 x 2 d y d x = 6 y − 3 x 2 3 y 2 − 6 x = 2 y − x 2 y 2 − 2 x . \frac{dy}{dx} = \frac{6y-3x^2}{3y^2-6x} = \frac{2y-x^2}{y^2-2x}. d x d y = 3 y 2 − 6 x 6 y − 3 x 2 = y 2 − 2 x 2 y − x 2 . ( 3 , 3 ) (3,3) ( 3 , 3 ) y ′ = 6 − 9 9 − 6 = − 3 3 = − 1 y'=\frac{6-9}{9-6}=\frac{-3}{3}=-1 y ′ = 9 − 6 6 − 9 = 3 − 3 = − 1 − 1 -1 − 1
Worked example Example 3 — Trig/transcendental:sin ( x y ) = x + y \sin(xy)=x+y sin ( x y ) = x + y cos ( x y ) ⋅ d d x ( x y ) = 1 + d y d x \cos(xy)\cdot\frac{d}{dx}(xy) = 1 + \frac{dy}{dx} cos ( x y ) ⋅ d x d ( x y ) = 1 + d x d y cos ( x y ) ( y + x d y d x ) = 1 + d y d x \cos(xy)\left(y + x\frac{dy}{dx}\right) = 1+\frac{dy}{dx} cos ( x y ) ( y + x d x d y ) = 1 + d x d y Why this step? Outer chain rule on sin \sin sin x y xy x y cannot solve for y y y only way.
Expand and collect:
y cos ( x y ) + x cos ( x y ) y ′ = 1 + y ′ y\cos(xy) + x\cos(xy)\,y' = 1 + y' y cos ( x y ) + x cos ( x y ) y ′ = 1 + y ′ y ′ ( x cos ( x y ) − 1 ) = 1 − y cos ( x y ) y'\big(x\cos(xy) - 1\big) = 1 - y\cos(xy) y ′ ( x cos ( x y ) − 1 ) = 1 − y cos ( x y ) d y d x = 1 − y cos ( x y ) x cos ( x y ) − 1 . \frac{dy}{dx} = \frac{1 - y\cos(xy)}{x\cos(xy) - 1}. d x d y = x c o s ( x y ) − 1 1 − y c o s ( x y ) .
Slope at ( a , b ) (a,b) ( a , b ) m = d y d x ∣ ( a , b ) m=\frac{dy}{dx}\big|_{(a,b)} m = d x d y ( a , b ) y − b = m ( x − a ) y-b=m(x-a) y − b = m ( x − a ) = − 1 / m =-1/m = − 1/ m
d d x ( arcsin x ) \frac{d}{dx}(\arcsin x) d x d ( arcsin x ) y = arcsin x ⇒ sin y = x y=\arcsin x \Rightarrow \sin y = x y = arcsin x ⇒ sin y = x cos y ⋅ d y d x = 1 ⇒ d y d x = 1 cos y . \cos y \cdot \frac{dy}{dx}=1 \;\Rightarrow\; \frac{dy}{dx}=\frac{1}{\cos y}. cos y ⋅ d x d y = 1 ⇒ d x d y = c o s y 1 . sin y = x \sin y = x sin y = x cos y = 1 − sin 2 y = 1 − x 2 \cos y=\sqrt{1-\sin^2 y}=\sqrt{1-x^2} cos y = 1 − sin 2 y = 1 − x 2 d d x arcsin x = 1 1 − x 2 . \frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1-x^2}}. d x d arcsin x = 1 − x 2 1 . This is how every inverse-function derivative is obtained.
Worked example Differentiatey = x x y = x^x y = x x ln \ln ln ln y = x ln x \ln y = x\ln x ln y = x ln x 1 y d y d x = ln x + 1 ⇒ d y d x = x x ( ln x + 1 ) . \frac{1}{y}\frac{dy}{dx} = \ln x + 1 \;\Rightarrow\; \frac{dy}{dx} = x^x(\ln x + 1). y 1 d x d y = ln x + 1 ⇒ d x d y = x x ( ln x + 1 ) . Why? You can't use the power rule (exponent is variable) or the exponential rule (base is variable). Logging turns the product/power into sums first.
Quantities linked by an equation, both changing in time t t t t t t x x x y y y t t t
d 2 y d x 2 \frac{d^2y}{dx^2} d x 2 d 2 y x 2 + y 2 = 25 x^2+y^2=25 x 2 + y 2 = 25 y ′ = − x / y y'=-x/y y ′ = − x / y y ′ = − x / y y'=-x/y y ′ = − x / y y ′ ′ = − ( 1 ) y − x y ′ y 2 = − y − x ( − x / y ) y 2 = − y + x 2 / y y 2 = − y 2 + x 2 y 3 = − 25 y 3 . y'' = -\frac{(1)y - x\,y'}{y^2} = -\frac{y - x(-x/y)}{y^2} = -\frac{y + x^2/y}{y^2} = -\frac{y^2+x^2}{y^3}=-\frac{25}{y^3}. y ′′ = − y 2 ( 1 ) y − x y ′ = − y 2 y − x ( − x / y ) = − y 2 y + x 2 / y = − y 3 y 2 + x 2 = − y 3 25 . y 3 y^3 y 3
Common mistake Forgetting thed y d x \frac{dy}{dx} d x d y Wrong: d d x ( y 2 ) = 2 y \frac{d}{dx}(y^2)=2y d x d ( y 2 ) = 2 y Why it feels right: it looks identical to d d x ( x 2 ) = 2 x \frac{d}{dx}(x^2)=2x d x d ( x 2 ) = 2 x Fix: y y y not the variable — it's a function of x x x d d x ( y 2 ) = 2 y d y d x \frac{d}{dx}(y^2)=2y\,\frac{dy}{dx} d x d ( y 2 ) = 2 y d x d y
Common mistake Using power rule instead of product rule onx y xy x y Wrong: d d x ( x y ) = y ′ \frac{d}{dx}(xy)=y' d x d ( x y ) = y ′ = 1 =1 = 1 Why it feels right: x y xy x y Fix: It's a product of two functions: d d x ( x y ) = 1 ⋅ y + x ⋅ y ′ = y + x y ′ \frac{d}{dx}(xy)=1\cdot y + x\cdot y' = y + xy' d x d ( x y ) = 1 ⋅ y + x ⋅ y ′ = y + x y ′
Common mistake Forgetting to substitute back before plugging in a pointWhen you differentiate again for y ′ ′ y'' y ′′ y ′ y' y ′ y ′ y' y ′
d d x ( 25 ) \frac{d}{dx}(25) d x d ( 25 ) 0 0 0 d d x \frac{d}{dx} d x d
Recall Feynman: explain to a 12-year-oldImagine x x x y y y x x x y y y exactly what y y y when x x x y y y " That ratio is d y d x \frac{dy}{dx} d x d y y y y y y y y y y x x x
Mnemonic Remember the toll"Touch a y y y y ′ y' y ′
Differentiate normally; every time you pass through a y y y d y d x \dfrac{dy}{dx} d x d y C.F.D. : C ollect, F actor, D ivide.
What does implicit differentiation assume about y y y That
y y y is a (differentiable) function of
x x x , even if we can't solve for it explicitly.
Why does d d x ( y 3 ) \frac{d}{dx}(y^3) d x d ( y 3 ) Chain rule —
y y y is inside, so
d d x ( y 3 ) = 3 y 2 d y d x \frac{d}{dx}(y^3)=3y^2\frac{dy}{dx} d x d ( y 3 ) = 3 y 2 d x d y .
d d x ( x y ) = ? \frac{d}{dx}(xy)=? d x d ( x y ) = ? y + x d y d x y + x\frac{dy}{dx} y + x d x d y (product rule).
For x 2 + y 2 = r 2 x^2+y^2=r^2 x 2 + y 2 = r 2 d y d x = ? \frac{dy}{dx}=? d x d y = ? General formula for slope of F ( x , y ) = 0 F(x,y)=0 F ( x , y ) = 0 d y d x = − F x F y \frac{dy}{dx}=-\frac{F_x}{F_y} d x d y = − F y F x .
Derive d d x arcsin x \frac{d}{dx}\arcsin x d x d arcsin x Set
sin y = x \sin y=x sin y = x , differentiate implicitly:
cos y y ′ = 1 \cos y\,y'=1 cos y y ′ = 1 , so
y ′ = 1 / 1 − x 2 y'=1/\sqrt{1-x^2} y ′ = 1/ 1 − x 2 .
d d x ( x x ) = ? \frac{d}{dx}(x^x)=? d x d ( x x ) = ? Logarithmic differentiation;
x x ( ln x + 1 ) x^x(\ln x+1) x x ( ln x + 1 ) .
The 4 steps of the technique? Differentiate both sides; add
y ′ y' y ′ to each
y y y -term; collect
y ′ y' y ′ terms; factor and divide.
When is implicit differentiation the only option? When the equation can't be solved for
y y y in closed form (e.g.
sin ( x y ) = x + y \sin(xy)=x+y sin ( x y ) = x + y ).
Second-derivative trap? Must substitute the expression for
y ′ y' y ′ back in; final answer shouldn't contain
y ′ y' y ′ .
Chain Rule — the engine behind every d y d x \frac{dy}{dx} d x d y Product Rule — needed for mixed x y xy x y Derivatives of Inverse Functions — derived via implicit differentiation.Logarithmic Differentiation — variable bases/exponents.Related Rates — implicit differentiation in time t t t Implicit Function Theorem — guarantees y ( x ) y(x) y ( x ) F y ≠ 0 F_y\neq 0 F y = 0 Tangent and Normal Lines — main geometric application.
guarantees where Fy not 0
Assume y is function of x
Implicit Function Theorem
Product rule for xy terms
Intuition Hinglish mein samjhoDekho, kabhi-kabhi equation aisa hota hai ki usme y y y x x x x 2 + y 2 = 25 x^2+y^2=25 x 2 + y 2 = 25 sin ( x y ) = x + y \sin(xy)=x+y sin ( x y ) = x + y y y y y = f ( x ) y=f(x) y = f ( x ) y y y y y y x x x x x x y y y d y d x \frac{dy}{dx} d x d y
Core funda: d d x ( x 2 ) = 2 x \frac{d}{dx}(x^2)=2x d x d ( x 2 ) = 2 x d d x ( y 2 ) = 2 y ⋅ d y d x \frac{d}{dx}(y^2)=2y\cdot\frac{dy}{dx} d x d ( y 2 ) = 2 y ⋅ d x d y y y y x x x x x x y y y x y xy x y d d x ( x y ) = y + x d y d x \frac{d}{dx}(xy)=y+x\frac{dy}{dx} d x d ( x y ) = y + x d x d y
Steps simple hain — Touch a y y y y ′ y' y ′ , phir C.F.D. (Collect, Factor, Divide): saare d y d x \frac{dy}{dx} d x d y d y d x \frac{dy}{dx} d x d y
Yeh technique kyun important hai? Kyunki isi se hum inverse functions ke derivatives nikaalte hain (jaise arcsin x ⇒ 1 / 1 − x 2 \arcsin x \Rightarrow 1/\sqrt{1-x^2} arcsin x ⇒ 1/ 1 − x 2 x x x^x x x
