Functions Defined by Integrals   

When functions are defined as arithmetic combinations of elementary functions, we don't really need the chain rule for functions of 2 variables in order to compute the derivative. For example, if w = x²y where x = t⁵ and y = t³, then simply substituting for x and y yields

w( t) = ( t5) 2·t3 = t13

and clearly, w' ( t) = 13t¹². However, many applications involve functions of two variables which are not elementary functions or are not stated explicitly. In such cases, the chain rule is essential.       

EXAMPLE 6    Evaluate the following derivative:

d dt ó õ t 0  sin( u2+t2) du

Solution: To begin with, we define

w = ó õ x 0  sin( u2+y) du

where x = t and y = t². The first partial with respect to x is

¶w ¶x =  ¶ ¶x ó õ x 0  sin( u2+y) du = sin( x2+y)

To find w_y, we will assume that the derivative with respect to y can be moved into the integrand (see exercise 35). Thus,

¶w ¶y =  ¶ ¶y ó õ x 0  sin( u2+y) du = ó õ x 0   ¶ ¶y sin( u2+y) du = ó õ x 0  cos( u2+y) du

Since x^¢( t) = 1 and y^¢( t) = 2t, the chain rule then yields

dw dt =  ¶w ¶x  dx dt +  ¶w ¶y  dy dt = sin( x2+y)  dx dt + æ è ó õ x 0  cos(u2+y) du ö ø     dy dt

Substituting x = t, y = t², x^¢ = 1, and y^¢ = 2t yields

w¢( t) = sin( 2t2) +2t ó õ t 0  cos( u2+t2) du

       

Moreover, differentiation of integrals is far from contrived. For example, in signal processing, the (Laplace) convolution of two functions f and g is defined

( f*g) ( t) = ó õ t 0  f( u) g(t-u) du

(5)

Convolution is the basis for the construction of analog and digital filters in electronics.       

EXAMPLE 7    What is the derivative of the convolution of the sinc function

f( t) =  sin( t) t ,  t ¹ 0

and the function g( t) = sin( t) .

       

Solution: According to (5), the convolution of f with g is

( f*g) ( t) = ó õ t 0   sin( u) u sin( t-u) du

Let us now let x = t, y = t, and

w = ó õ x 0   sin( u) u sin( y-u) du

The partial derivatives of w are

¶w ¶x =  ¶ ¶x ó õ x 0   sin( u) u sin( y-u) du =  sin(x) x sin( y-x)  ¶w ¶y =  ¶ ¶y ó õ x 0   sin( u) u sin( y-u) du = ó õ x 0   sin( u) u  ¶ ¶y sin( y-u)du = ó õ x 0   sin( u) u cos( y-u) du

Since dx/dt = dy/dt = 1, the chain rule reduces to

dw dt =  ¶w ¶x  dx dt +  ¶w ¶y  dy dt =  sin( x) x sin(y-x) + ó õ x 0   sin( u) u cos(y-u) du