6. (10 Points) Can one claim that (condz)(x)(condf)(x)+(condg)(x)+(condh)(x)+(condj)(x)+(condk)(x) where z(x)=f(x)g(x)h(x)j(x)k(x) ? You may genaralize the problem to n functions. (a) Start by computing z(x)=dxd(f(x)g(x)h(x)j(x)k(x)). (b) Compute z(x)z(x) and simplify the result. (c) Use the general form of the condition number. (d) State your final