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et f x y t 1 y f x t d t for x x 1 x 2 and y t 1 t 2 displaystyle f x y int _ t_ 1 y f x t dt qquad text for x in x_ 1 x_ 2 text and y in t_ 1 t_ 2 and g x a x b x f x t d t for x x 1 x 2 displaystyle g x int _ a x b x f x t dt quad text for x in x_ 1 x_ 2 then by properties of definite integrals we can write g x t 1 b x f x t d t t 1 a x f x t d t f x b x f x a x displaystyle g x int _ t_ 1 b x f x t dt int _ t_ 1 a x f x t dt f x b x f x a x since the functions f a b displaystyle f a b are all differentiable see the remark at the end of the proof by the multivariable chain rule it follows that g displaystyle g is differentiable and its derivative is given by the formula g x f x x b x f b x x b x b x f x x a x f a x x a x a x displaystyle g x left frac partial f partial x x b x frac partial f partial b x x b x b x right left frac partial f partial x x a x frac partial f partial a x x a x a x right now note that for every x x 1 x 2 displaystyle x in x_ 1 x_ 2 and for every y t 1 t 2 displaystyle y in t_ 1 t_ 2 we have that f x x y t 1 y f x x t d t textstyle frac partial f partial x x y int _ t_ 1 y frac partial f partial x x t dt because when taking the partial derivative with respect to x displaystyle x of f displaystyle f we are keeping y displaystyle y fixed in the expression t 1 y f x t d t textstyle int _ t_ 1 y f x t dt thus the basic form of leibniz s integral rule with constant limits of integration applies next by the first fundamental theorem of calculus we have that f y x y f x y textstyle frac partial f partial y x y f x y because when taking the partial derivative with respect to y displaystyle y of f displaystyle f the first variable x displaystyle x is fixed so the fundamental theorem can indeed be applied substituting these results into the equation for g x displaystyle g x above gives g x t 1 b x f x x t d t f x b x b x t 1 a x f x x t d t f x a x a x f x b x b x f x a x a x a x b x f x x t d t displaystyle begin aligned g x left int _ t_ 1 b x frac partial f partial x x t dt f x b x b x right left int _ t_ 1 a x dfrac partial f partial x x t dt f x a x a x right 2pt f x b x b x f x a x a x int _ a x b x frac partial f partial x x t dt end aligned as desired there is a technical point in the proof above which is worth noting applying the chain rule to g displaystyle g requires that f displaystyle f already be differentiable this is where we use our assumptions about f displaystyle f as mentioned above the partial derivatives of f displaystyle f are given by the formulas f x x y t 1 y f x x t d t textstyle frac partial f partial x x y int _ t_ 1 y frac partial f partial x x t dt and f y x y f x y textstyle frac partial f partial y x y f x y since f x textstyle dfrac partial f partial x is continuous its integral is also a continuous function 8 and since f displaystyle f is also continuous these two results show that both the partial derivatives of f displaystyle f are continuous since continuity of partial derivatives implies differentiability of the function 9 f displaystyle f is indeed differentiable three dimensional time dependent form edit see also higher dimensions at time t the surface σ in figure 1 contains a set of points arranged about a centroid c t displaystyle mathbf c t the function f r t displaystyle mathbf f mathbf r t can be written as f c t r c t t f c t i t displaystyle mathbf f mathbf c t mathbf r mathbf c t t mathbf f mathbf c t mathbf i t with i displaystyle mathbf i independent of time variables are shifted to a new frame of reference attached to the moving surface with origin at c t displaystyle mathbf c t for a rigidly translating surface the limits of integration are then independent of time so d d t σ t d a r f r t σ d a i d d t f c t i t displaystyle frac d dt left iint _ sigma t d mathbf a _ mathbf r cdot mathbf f mathbf r t right iint _ sigma d mathbf a _ mathbf i cdot frac d dt mathbf f mathbf c t mathbf i t where the limits of integration confining the integral to the region σ no longer are time dependent so differentiation passes through the integration to act on the integrand only d d t f c t i t f t c t i t v f c t i t f t r t v f r t displaystyle frac d dt mathbf f mathbf c t mathbf i t mathbf f _ t mathbf c t mathbf i t mathbf v cdot nabla f mathbf c t mathbf i t mathbf f _ t mathbf r t mathbf v cdot nabla mathbf f mathbf r t with the velocity of motion of the surface defined by v d d t c t displaystyle mathbf v frac d dt mathbf c t this equation expresses the material derivative of the field that is the derivative with respect to a coordinate system attached to the moving surface having found the derivative variables can be switched back to the original frame of reference we notice that see article on curl v f f f v v v f displaystyle nabla times left mathbf v times mathbf f right nabla cdot mathbf f mathbf f cdot nabla mathbf v nabla cdot mathbf v mathbf v cdot nabla mathbf f and that stokes theorem equates the surface integral of the curl over σ with a line integral over σ d d t σ t f r t d a σ t f t r t f v f v v f d a σ t v f d s displaystyle frac d dt left iint _ sigma t mathbf f mathbf r t cdot d mathbf a right iint _ sigma t big mathbf f _ t mathbf r t left mathbf f cdot nabla right mathbf v left nabla cdot mathbf f right mathbf v nabla cdot mathbf v mathbf f big cdot d mathbf a oint _ partial sigma t left mathbf v times mathbf f right cdot d mathbf s the sign of the line integral is based on the right hand rule for the choice of direction of line element d s to establish this sign for example suppose the field f points in the positive z direction and the surface σ is a portion of the xy plane with perimeter σ we adopt the normal to σ to be in the positive z direction positive traversal of σ is then counterclockwise right hand rule with thumb along z axis then the integral on the left hand side determines a positive flux of f through σ suppose σ translates in the positive x direction at velocity v an element of the boundary of σ parallel to the y axis say d s sweeps out an area v t d s in time t if we integrate around the boundary σ in a counterclockwise sense v t d s points in the negative z direction on the left side of σ where d s points downward and in the positive z direction on the right side of σ where d s points upward which makes sense because σ is moving to the right adding area on the right and losing it on the left on that basis the flux of f is increasing on the right of σ and decreasing on the left however the dot product v f d s f v d s f v d s consequently the sign of the line integral is taken as negative if v is a constant d d t σ t f r t d a σ t f t r t f v d a σ t v f d s displaystyle frac d dt iint _ sigma t mathbf f mathbf r t cdot d mathbf a iint _ sigma t big mathbf f _ t mathbf r t left nabla cdot mathbf f right mathbf v big cdot d mathbf a oint _ partial sigma t left mathbf v times mathbf f right cdot d mathbf s which is the quoted result this proof does not consider the possibility of the surface deforming as it moves alternative derivation edit lemma one has b a b f x d x f b a a b f x d x f a displaystyle frac partial partial b left int _ a b f x dx right f b qquad frac partial partial a left int _ a b f x dx right f a proof from the proof of the fundamental theorem of calculus b a b f x d x lim δ b 0 1 δ b a b δ b f x d x a b f x d x lim δ b 0 1 δ b a b f x d x b b δ b f x d x a b f x d x lim δ b 0 1 δ b b b δ b f x d x lim δ b 0 1 δ b f b δ b o δ b 2 f b displaystyle begin aligned frac partial partial b left int _ a b f x dx right lim _ delta b to 0 frac 1 delta b left int _ a b delta b f x dx int _ a b f x dx right 1ex lim _ delta b to 0 frac 1 delta b left int _ a b f x dx int _ b b delta b f x dx int _ a b f x dx right 1ex lim _ delta b to 0 frac 1 delta b int _ b b delta b f x dx 1ex lim _ delta b to 0 frac 1 delta b left f b delta b o left delta b 2 right right 1ex f b end aligned and a a b f x d x lim δ a 0 1 δ a a δ a b f x d x a b f x d x lim δ a 0 1 δ a a δ a a f x d x lim δ a 0 1 δ a f a δ a o δ a 2 f a displaystyle begin aligned frac partial partial a left int _ a b f x dx right lim _ delta a to 0 frac 1 delta a left int _ a delta a b f x dx int _ a b f x dx right 6pt lim _ delta a to 0 frac 1 delta a int _ a delta a a f x dx 6pt lim _ delta a to 0 frac 1 delta a left f a delta a o left delta a 2 right right 6pt f a end aligned suppose a and b are constant and that f x involves a parameter α which is constant in the integration but may vary to form different integrals assume that f x α is a continuous function of x and α in the compact set x α α 0 α α 1 and a x b and that the partial derivative f α x α exists and is continuous if one defines φ α a b f x α d x displaystyle varphi alpha int _ a b f x alpha dx then φ displaystyle varphi may be differentiated with respect to α by differentiating under the integral sign i e d φ d α a b α f x α d x displaystyle frac d varphi d alpha int _ a b frac partial partial alpha f x alpha dx by the heine cantor theorem it is uniformly continuous in that set in other words for any ε 0 there exists δ α such that for all values of x in a b f x α δ α f x α ε displaystyle f x alpha delta alpha f x alpha varepsilon on the other hand δ φ φ α δ α φ α a b f x α δ α d x a b f x α d x a b f x α δ α f x α d x ε b a displaystyle begin aligned delta varphi varphi alpha delta alpha varphi alpha 6pt int _ a b f x alpha delta alpha dx int _ a b f x alpha dx 6pt int _ a b left f x alpha delta alpha f x alpha right dx 6pt leq varepsilon b a end aligned hence φ α is a continuous function similarly if α f x α displaystyle frac partial partial alpha f x alpha exists and is continuous then for all ε 0 there exists δ α such that x a b f x α δ α f x α δ α f α ε displaystyle forall x in a b quad left frac f x alpha delta alpha f x alpha delta alpha frac partial f partial alpha right varepsilon therefore δ φ δ α a b f x α δ α f x α δ α d x a b f x α α d x r displaystyle frac delta varphi delta alpha int _ a b frac f x alpha delta alpha f x alpha delta alpha dx int _ a b frac partial f x alpha partial alpha dx r where r a b ε d x ε b a displaystyle r int _ a b varepsilon dx varepsilon b a now ε 0 as δ α 0 so lim δ α 0 δ φ δ α d φ d α a b α f x α d x displaystyle lim _ delta alpha to 0 frac delta varphi delta alpha frac d varphi d alpha int _ a b frac partial partial alpha f x alpha dx this is the formula we set out to prove now suppose a b f x α d x φ α displaystyle int _ a b f x alpha dx varphi alpha where a and b are functions of α which take increments δ a and δ b respectively when α is increased by δ α then δ φ φ α δ α φ α a δ a b δ b f x α δ α d x a b f x α d x a δ a a f x α δ α d x a b f x α δ α d x b b δ b f x α δ α d x a b f x α d x a a δ a f x α δ α d x a b f x α δ α f x α d x b b δ b f x α δ α d x displaystyle begin aligned delta varphi varphi alpha delta alpha varphi alpha 6pt int _ a delta a b delta b f x alpha delta alpha dx int _ a b f x alpha dx 6pt int _ a delta a a f x alpha delta alpha dx int _ a b f x alpha delta alpha dx int _ b b delta b f x alpha delta alpha dx int _ a b f x alpha dx 6pt int _ a a delta a f x alpha delta alpha dx int _ a b f x alpha delta alpha f x alpha dx int _ b b delta b f x alpha delta alpha dx end aligned a form of the mean value theorem a b f x d x b a f ξ textstyle int _ a b f x dx b a f xi where a ξ b can be applied to the first and last integrals of the formula for δ φ above resulting in δ φ δ a f ξ 1 α δ α a b f x α δ α f x α d x δ b f ξ 2 α δ α displaystyle delta varphi delta a f xi _ 1 alpha delta alpha int _ a b f x alpha delta alpha f x alpha dx delta b f xi _ 2 alpha delta alpha dividing by δ α letting δ α 0 noticing ξ 1 a and ξ 2 b and using the above derivation for d φ d α a b α f x α d x displaystyle frac d varphi d alpha int _ a b frac partial partial alpha f x alpha dx yields d φ d α a b α f x α d x f b α b α f a α a α displaystyle frac d varphi d alpha int _ a b frac partial partial alpha f x alpha dx f b alpha frac partial b partial alpha f a alpha frac partial a partial alpha this is the general form of the leibniz integral rule examples edit example 1 fixed limits edit consider the function φ α 0 1 α x 2 α 2 d x displaystyle varphi alpha int _ 0 1 frac alpha x 2 alpha 2 dx the function under the integral sign is not continuous at the point x α 0 0 displaystyle x alpha 0 0 and the function φ α displaystyle varphi alpha has a discontinuity at α 0 displaystyle alpha 0 because φ α displaystyle varphi alpha approaches π 2 displaystyle pm pi 2 as α 0 displaystyle alpha to 0 pm if we differentiate φ α displaystyle varphi alpha with respect to α displaystyle alpha under the integral sign we get d d α φ α 0 1 α α x 2 α 2 d x 0 1 x 2 α 2 x 2 α 2 2 d x x x 2 α 2 0 1 1 1 α 2 displaystyle frac d d alpha varphi alpha int _ 0 1 frac partial partial alpha left frac alpha x 2 alpha 2 right dx int _ 0 1 frac x 2 alpha 2 x 2 alpha 2 2 dx left frac x x 2 alpha 2 right _ 0 1 frac 1 1 alpha 2 for α 0 displaystyle alpha neq 0 this may be integrated with respect to α displaystyle alpha to find φ α 0 α 0 arctan α π 2 α 0 displaystyle varphi alpha begin cases 0 alpha 0 arctan alpha frac pi 2 alpha neq 0 end cases example 2 variable limits edit an example with variable limits d d x sin x cos x cosh t 2 d t cosh cos 2 x d d x cos x cosh sin 2 x d d x sin x sin x cos x x cosh t 2 d t cosh cos 2 x sin x cosh sin 2 x cos x 0 cosh cos 2 x sin x cosh sin 2 x cos x displaystyle begin aligned frac d dx int _ sin x cos x cosh t 2 dt cosh left cos 2 x right frac d dx cos x cosh left sin 2 x right frac d dx sin x int _ sin x cos x frac partial partial x cosh t 2 dt 6pt cosh cos 2 x sin x cosh sin 2 x cos x 0 6pt cosh cos 2 x sin x cosh sin 2 x cos x end aligned applications edit evaluating definite integrals edit the formula d d x a x b x f x t d t f x b x d d x b x f x a x d d x a x a x b x x f x t d t displaystyle frac d dx left int _ a x b x f x t dt right f big x b x big cdot frac d dx b x f big x a x big cdot frac d dx a x int _ a x b x frac partial partial x f x t dt can be of use when evaluating certain definite integrals when used in this context the leibniz integral rule for differentiating under the integral sign is also known as feynman s trick for integration example 3 edit consider φ α 0 π ln 1 2 α cos x α 2 d x α 1 displaystyle varphi alpha int _ 0 pi ln left 1 2 alpha cos x alpha 2 right dx qquad alpha neq 1 its derivative known as the poisson integral 10 is given by d d α φ α 0 π 2 cos x 2 α 1 2 α cos x α 2 d x 1 α 0 π 1 1 α 2 1 2 α cos x α 2 d x π α 2 α arctan 1 α 1 α tan x 2 0 π displaystyle begin aligned frac d d alpha varphi alpha int _ 0 pi frac 2 cos x 2 alpha 1 2 alpha cos x alpha 2 dx 6pt frac 1 alpha int _ 0 pi left 1 frac 1 alpha 2 1 2 alpha cos x alpha 2 right dx 6pt left frac pi alpha frac 2 alpha left arctan left frac 1 alpha 1 alpha tan left frac x 2 right right right right _ 0 pi end aligned as x displaystyle x varies from 0 displaystyle 0 to π d...