Derivative Rules — Complete Calculus Reference A complete reference of derivative rules and formulas — basic rules, special functions, and inverse derivatives. Use this as a calculus cheat sheet for homework, exams, and quick checks. Each formula links to its definition page where available.
Basic Differentiation Rules Constant Rule
d d x [ c ] = 0 \frac{d}{dx}[c] = 0 d x d [ c ] = 0 Power Rule
d d x [ x n ] = n x n − 1 \frac{d}{dx}[x^n] = n x^{n-1} d x d [ x n ] = n x n − 1 Constant Multiple
d d x [ c f ( x ) ] = c f ′ ( x ) \frac{d}{dx}[c\,f(x)] = c\,f'(x) d x d [ c f ( x )] = c f ′ ( x ) Sum Rule
d d x [ f ( x ) + g ( x ) ] = f ′ ( x ) + g ′ ( x ) \frac{d}{dx}[f(x) + g(x)] = f'(x) + g'(x) d x d [ f ( x ) + g ( x )] = f ′ ( x ) + g ′ ( x ) Difference Rule
d d x [ f ( x ) − g ( x ) ] = f ′ ( x ) − g ′ ( x ) \frac{d}{dx}[f(x) - g(x)] = f'(x) - g'(x) d x d [ f ( x ) − g ( x )] = f ′ ( x ) − g ′ ( x ) Product Rule
d d x [ f g ] = f ′ g + f g ′ \frac{d}{dx}[f g] = f' g + f g' d x d [ f g ] = f ′ g + f g ′ Quotient Rule
d d x [ f g ] = f ′ g − f g ′ g 2 \frac{d}{dx}\!\left[\frac{f}{g}\right] = \frac{f' g - f g'}{g^2} d x d [ g f ] = g 2 f ′ g − f g ′ Chain Rule
d d x [ f ( g ( x ) ) ] = f ′ ( g ( x ) ) ⋅ g ′ ( x ) \frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x) d x d [ f ( g ( x ))] = f ′ ( g ( x )) ⋅ g ′ ( x ) Exponential & Logarithmic Derivatives Natural Exponential
d d x [ e x ] = e x \frac{d}{dx}[e^x] = e^x d x d [ e x ] = e x General Exponential
d d x [ a x ] = a x ln a \frac{d}{dx}[a^x] = a^x \ln a d x d [ a x ] = a x ln a Exponential (Chain)
d d x [ e u ( x ) ] = e u ( x ) ⋅ u ′ ( x ) \frac{d}{dx}[e^{u(x)}] = e^{u(x)} \cdot u'(x) d x d [ e u ( x ) ] = e u ( x ) ⋅ u ′ ( x ) Natural Logarithm
d d x [ ln x ] = 1 x \frac{d}{dx}[\ln x] = \frac{1}{x} d x d [ ln x ] = x 1 Logarithm Base a
d d x [ log a x ] = 1 x ln a \frac{d}{dx}[\log_a x] = \frac{1}{x \ln a} d x d [ log a x ] = x ln a 1 Log (Chain)
d d x [ ln u ( x ) ] = u ′ ( x ) u ( x ) \frac{d}{dx}[\ln u(x)] = \frac{u'(x)}{u(x)} d x d [ ln u ( x )] = u ( x ) u ′ ( x ) Trigonometric Derivatives Sine
d d x [ sin x ] = cos x \frac{d}{dx}[\sin x] = \cos x d x d [ sin x ] = cos x Cosine
d d x [ cos x ] = − sin x \frac{d}{dx}[\cos x] = -\sin x d x d [ cos x ] = − sin x Tangent
d d x [ tan x ] = sec 2 x \frac{d}{dx}[\tan x] = \sec^2 x d x d [ tan x ] = sec 2 x Cotangent
d d x [ cot x ] = − csc 2 x \frac{d}{dx}[\cot x] = -\csc^2 x d x d [ cot x ] = − csc 2 x Secant
d d x [ sec x ] = sec x tan x \frac{d}{dx}[\sec x] = \sec x \tan x d x d [ sec x ] = sec x tan x Cosecant
d d x [ csc x ] = − csc x cot x \frac{d}{dx}[\csc x] = -\csc x \cot x d x d [ csc x ] = − csc x cot x Inverse Trigonometric Derivatives Arcsine
d d x [ sin − 1 x ] = 1 1 − x 2 \frac{d}{dx}[\sin^{-1} x] = \frac{1}{\sqrt{1-x^2}} d x d [ sin − 1 x ] = 1 − x 2 1 Arccosine
d d x [ cos − 1 x ] = − 1 1 − x 2 \frac{d}{dx}[\cos^{-1} x] = -\frac{1}{\sqrt{1-x^2}} d x d [ cos − 1 x ] = − 1 − x 2 1 Arctangent
d d x [ tan − 1 x ] = 1 1 + x 2 \frac{d}{dx}[\tan^{-1} x] = \frac{1}{1+x^2} d x d [ tan − 1 x ] = 1 + x 2 1 Arccotangent
d d x [ cot − 1 x ] = − 1 1 + x 2 \frac{d}{dx}[\cot^{-1} x] = -\frac{1}{1+x^2} d x d [ cot − 1 x ] = − 1 + x 2 1 Arcsecant
d d x [ sec − 1 x ] = 1 ∣ x ∣ x 2 − 1 \frac{d}{dx}[\sec^{-1} x] = \frac{1}{|x|\sqrt{x^2-1}} d x d [ sec − 1 x ] = ∣ x ∣ x 2 − 1 1 Arccosecant
d d x [ csc − 1 x ] = − 1 ∣ x ∣ x 2 − 1 \frac{d}{dx}[\csc^{-1} x] = -\frac{1}{|x|\sqrt{x^2-1}} d x d [ csc − 1 x ] = − ∣ x ∣ x 2 − 1 1 Hyperbolic Derivatives Hyperbolic Sine
d d x [ sinh x ] = cosh x \frac{d}{dx}[\sinh x] = \cosh x d x d [ sinh x ] = cosh x Hyperbolic Cosine
d d x [ cosh x ] = sinh x \frac{d}{dx}[\cosh x] = \sinh x d x d [ cosh x ] = sinh x Hyperbolic Tangent
d d x [ tanh x ] = sech 2 x \frac{d}{dx}[\tanh x] = \operatorname{sech}^2 x d x d [ tanh x ] = sech 2 x Hyperbolic Secant
d d x [ sech x ] = − sech x tanh x \frac{d}{dx}[\operatorname{sech} x] = -\operatorname{sech} x \tanh x d x d [ sech x ] = − sech x tanh x Special Techniques Implicit Differentiation
d d x [ F ( x , y ) ] = 0 ⟹ solve for d y d x \frac{d}{dx}[F(x,y)] = 0 \implies \text{solve for } \tfrac{dy}{dx} d x d [ F ( x , y )] = 0 ⟹ solve for d x d y Logarithmic Differentiation
y = f ( x ) g ( x ) ⟹ ln y = g ( x ) ln f ( x ) y = f(x)^{g(x)} \implies \ln y = g(x) \ln f(x) y = f ( x ) g ( x ) ⟹ ln y = g ( x ) ln f ( x ) Inverse Function Theorem
( f − 1 ) ′ ( y ) = 1 f ′ ( f − 1 ( y ) ) (f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))} ( f − 1 ) ′ ( y ) = f ′ ( f − 1 ( y )) 1 Parametric Derivative
d y d x = d y / d t d x / d t \frac{dy}{dx} = \frac{dy/dt}{dx/dt} d x d y = d x / d t d y / d t Second Derivative (Parametric)
d 2 y d x 2 = d d t ( d y d x ) d x / d t \frac{d^2 y}{dx^2} = \frac{\frac{d}{dt}\!\left(\frac{dy}{dx}\right)}{dx/dt} d x 2 d 2 y = d x / d t d t d ( d x d y ) 