Hoofdstuk 12 Differentiëren *: Som- en productregel *
The derivative of a sum function
Let #f# and #g# be two functions, and #a# and #b# two numbers.
The sum function #af+bg# is known from Operators on functions. It is the function that assigns the value #(af+bg)(x) = a\cdot f(x) + b\cdot g(x)# to #x#.
Hence, #af# is the function that assigns the value #a\cdot f(x)# to #x: (af)(x) = a\cdot f(x)#.
If, for example, #a =3# and #f(x) = x^2+1#, then the function #af# assigns the value #a\cdot f(x) = 3(x^2+1)=3x^2+3# to #x#.
The sum Rule for Differentiation
The derivative of #af+bg# is #af'+bg'#.
Because, the difference quotient of #af+bg# is equal to
\[\begin{array}{rcl}\dfrac{(af+bg)(x) -(af+bg)(c)}{x-c} &=& \dfrac{af(x)+bg(x)-af(c)-bg(c)}{x-c}\\ & =& a\dfrac{f(x)-f(c)}{x-c} + b\dfrac{g(x)-g(c)}{x-c} \tiny.\end{array}\]
From this the derivative follows:
\[\begin{array}{rcl}(af+bg)'(c)&=&\lim_{x\to c}\left(a\dfrac{f(x)-f(c)}{x-c} + b\dfrac{g(x)-g(c)}{x-c}\right)\\ & =& a\lim_{x\to c} \dfrac{f(x)-f(c)}{x-c} + b\lim_{x\to c}\dfrac{g(x)-g(c)}{x-c}\\ &=& af'(c)+bg'(c).\end{array}\]
Consequently, #(af+bg)'(c)= af'(c)+bg'(c)#.
Because, the requirements are #(f+l)(0) = 0# and #(f+l)'(1) = 0#. The first equation can be rewritten as #f(0)+l(0) = 0# and gives #10+b = 0#, so #b=-10#. By using the sum rule for differentiation, the second equation can be rewritten as #f'(1)+l'(1) = 0# and gives #4+a=0#, so #a = -4#. Therefore, #l(x) = -4x-10#.
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