Leibniz’ Rule - University of Texas at Dallas

Leibniz' Rule

d v(x)

dv

du v(x) f (x, t)

f (x, t)dt = f (x, v(x)) - f (x, u(x)) +

dt

dx u(x)

dx

dx u(x) x

1

d 2/x sin xt dt

dx t=1/x t

sin 2 2 sin 1 1

2/x

= ?- - ?- +

cos xt dt

2/x x2 1/x x2 t=1/x

sin 2 sin 1 sin xt 2/x

=- + +

x

x

x t=1/x

sin 2 sin 1 sin 2 sin 1

=- + +

-

=0

x

x

x

x

2

v 1 - e-t

s=

dt

ut

s 1 - ev l H -ev

=

-1

v v 1

s 1 - eu l H +eu

=-

+1

u

u 1

3

Evaluate the integral

1 x - 1

F () =

dx ( 0)

0 ln x

by differentiating under the integral sign

Differentiate both sides with respect to :

1 x - 1

F () =

dx

0 ln x

=

1 1 x ln x dx =

1

x dx =

x+1 1 =

1

0 ln x

0

+1 0 1+

Integrating now with respect to we obtain F () = ln(1 + ) + C.

Since F (0) = 0, C = 0 F () = ln(1 + )

4

Evaluate the integral

F () =

e-x ?

sin x

dx

0

x

Differentiate both sides with respect to :

5

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