Question 375478
{{{ y = (x^2 - 1)/(x^2 + 2x + 1) }}}
{{{y=(x^2-1)/(x+1)^2}}}
Use the quotient rule,
{{{ dy/dx = ((x+1)^2(2x)-(x^2-1)(2(x+1)))/(x+1)^4}}}
{{{dy/dx=(2x(x+1)-2(x^2-1))/(x+1)^3}}}
{{{dy/dx=(2x^2+2x-2x^2+2))/(x+1)^3}}}
{{{dy/dx=(2(x+1))/(x+1)^3}}}
{{{dy/dx=2/(x+1)^2}}}
When {{{x=0}}},
{{{dy/dx=2/(0+1)^2}}}
{{{highlight(dy/dx=2)}}}
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{{{y=(e^x + 1)/(1 + e^(-x))}}}
{{{dy/dx=((1+e^(-x))(e^(x))-(e^x+1)(-e^(-x)))/(1+e^(-x))^2}}}
 {{{dy/dx=((e^(x)+1+1+e^(-x)))/(1+e^(-x))^2}}}
 {{{dy/dx=((e^(x)+e^(-x)+2))/(1+e^(-x))^2}}}
 {{{dy/dx=((e^(x)+e^(-x)+2))/(1+e^(-x))^2}}}
{{{dy/dx=((e^(x)+e^(-x)+2))/(e^(-2x)+2e^(-x)+1)}}}
 {{{dy/dx=e^(x)}}}
When {{{x=0}}},
{{{dy/dx=e^(0)}}}
{{{highlight(dy/dx=1)}}}