解微積分兩題

1.
Let y=(x+1/x)(x-1/x+1).
Find y'
(a) by applying the product rule and
(b) by multiplying the factor to produce a sum of simpler terms to differentiate.

(a) Product rule: (fg)'=f'g+fg'
let
f(x)=(x+1/x)
f'(x)=1-/x^2)
g(x)=(x-1/x+1)
g'(x)=1+1/x^2
f'
y=fg
y'(x)=f'g+g'f
=(1-1/x^2)(x-1/x+1)+(x+1/x)(1+1/x^2)
=(2x^4+x^3-x+2)/x^3
=2x-1/x^2+2/x^3

(b) simplify before differentiation
y(x)=(x+1/x)(x-1/x+1)
=x^2+x+1/x-1/x^2
y'(x)=2x-1/x^2+2/x^3

2.
The curves y1(x)=x^2+ax+b and y2(x)=cx-x^2 have a common tangent line at the point (1,0).
Find a,b and c.

y1(1)=1+a+b=0
=> a+b+1=0
=>y1(x)=x^2+ax-(a+1) by substituting b=-a-1
y1'(x)=2x+a

y2(1)=c-1=0
=>c=1
=> y2(x)=x-x^2
y2'(x)=1-2x

At (1,0), they have a common tangent, therefore
y1'(1)=y2'(1)
=> 2+a=1-2=-1
=> a=-1-2=-3

from :1+a+b=0
b=-1-a=2

Thus, substituting into y1(x) and y2(x),
y1(x)=x^2-3x+2
y2(x)=x-x^2

check:
y1(1)=1-3+2=0; y1'(x)=2x+a=2-3=-1
y2(1)=1-1=0; y2'(x)=1-2x=1-2=-1