Question

How do you find the slope of a tangent line to the graph of the function `f(x) = x^2/(x+1)` at (2, 4/3)?

Answer 1

`"slope "=8/9`

Explanation:

`color(orange)"Reminder"`

The slope of the tangent at x = a on f(x) is f'(a)

To differentiate f(x) use the `color(blue)"quotient rule"`

`"given " f(x)=(g(x))/(h(x))" then "`

`color(red)(bar(ul(|color(white)(2/2)color(black)(f'(x)=(h(x)g'(x)-g(x)h'(x))/(h(x))^2)color(white)(2/2)|)))`

`"here " g(x)=x^2rArrg'(x)=2x`

`"and " h(x)=x+1rArrh'(x)=1`

`rArrf'(x)=((x+1).2x-x^2 .1)/(x+1)^2=(2x^2+2x-x^2)/(x+1)^2`

`=(x^2+2x)/(x+1)^2`

`rArrf'(2)=(4+4)/3^2=8/9`