Difference Quotient Generalized Equations

[Eigenvalue]

y=ax+b
difference quotient:[tex]\frac{a(x+h)+b-ax-b}{h}[/tex]

=[tex]\frac{ax+ah-ax}{h}[/tex]

=[tex]\frac{ah}{h}[/tex]

=a


y=ax²+bx+c
difference quotient:[tex]\frac{a(x+h)²+b(x+h)+c-ax²-bx-c}{h}[/tex]

=[tex]\frac{2ahx+ah²+bh}{h}[/tex]

=2ax+ah+b

y=a^x
difference quotient: [tex]\frac{ a^{x+h }- a^{x } }{h}[/tex]

=[tex]\frac{ a^{x }( a^{h }-1) }{h}[/tex]

[Eigenvalue]

y=sin(x)
difference quotient: [tex]\frac{sin(x+h)-sin(x)}{h}[/tex]
use angle addition identity for sine
=[tex]\frac{(sin(x)cos(h)+sin(h)cos(x))-sin(x)}{h}[/tex]

=[tex]\frac{sin(x)(cos(x)-1)+cos(x)sin(h)}{h}[/tex]

[Eigenvalue]

y=cos(x)
difference quotient: [tex]\frac{cos(x+h)-cos(x)}{h}[/tex]
use angle sum identity for cosine
=[tex]\frac{(cos(x)cos(h)-sin(x)sin(h))-cos(x)}{h}[/tex]

=[tex]\frac{cos(x)(cos(h)-1)-sin(x)sin(h)}{h}[/tex]

[Eigenvalue]

y=tan(x)
difference quotient: [tex]\frac{tan(x+h)-tan(x)}{h}[/tex]
Use the angle sum identity for tangent

=[tex]\frac{tan(x)+tan(h)-tan(x)(1-tan(x)tan(h))}{h(1-(tan(x)tan(h))}[/tex]

=[tex]\frac{tan(h)(1+tan²(x))}{h(1-tan(x)tan(h))}[/tex]

=[tex]\frac{tan(h)sec²(x)}{h(1-tan(x)tan(h))}[/tex]

[Eigenvalue]

y=[tex]\frac{ax+b}{cx+d}[/tex]

difference quotient:[tex]\frac{(a(x+h)+b)/(c(x+h)+d)-((ax+b)/(cx+d))}{h}[/tex]

=[tex]\frac{(ax+ah+b)(cx+d)-(cx+ch+d)(ax+b)}{h}[/tex]

=[tex]\frac{h(ad-bc)}{h(cx+ch+d)(cx+d)}[/tex]

=[tex]\frac{ad-bc}{(cx+ch+d)(cx+d)}[/tex]

[nycmath]

y=ax+b
difference quotient:[tex]\frac{a(x+h)+b-ax-b}{h}[/tex]

=[tex]\frac{ax+ah-ax}{h}[/tex]

=[tex]\frac{ah}{h}[/tex]

=a


y=ax²+bx+c
difference quotient:[tex]\frac{a(x+h)²+b(x+h)+c-ax²-bx-c}{h}[/tex]

=[tex]\frac{2ahx+ah²+bh}{h}[/tex]

=2ax+ah+b

y=a^x
difference quotient: [tex]\frac{ a^{x+h }- a^{x } }{h}[/tex]

=[tex]\frac{ a^{x }( a^{h }-1) }{h}[/tex]

I enjoy breaking down questions leading to a generalized equation or formula.

[nycmath]

y=[tex]\frac{ax+b}{cx+d}[/tex]

difference quotient:[tex]\frac{(a(x+h)+b)/(c(x+h)+d)-((ax+b)/(cx+d))}{h}[/tex]

=[tex]\frac{(ax+ah+b)(cx+d)-(cx+ch+d)(ax+b)}{h}[/tex]

=[tex]\frac{h(ad-bc)}{h(cx+ch+d)(cx+d)}[/tex]

=[tex]\frac{ad-bc}{(cx+ch+d)(cx+d)}[/tex]

Another great, detailed and mathematical reply. Thank you. Some information about the difference quotient of f and questions from the David Cohen book coming up.