How do you find the slope of the tangent line to the graph of $f (x) = \frac{x^{2}}{x + 1}$ $f(x)=x^2/(x+1)$ at (2, 4/3)?

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Answer 1 · 2015-05-08T17:38:19

$\frac{8}{9}$ $8/9$

Find f'(x) = $\frac{2 x (x + 1) - x^{2} \cdot 1}{{(x + 1)}^{2}}$ $(2x(x+1)-x^2 *1)/(x+1)^2$

= $\frac{x^{2} + 2 x}{{(x + 1)}^{2}}$ $(x^2+2x)/((x+1)^2$

At x=2, f'(x)= 8/9

How do you find the slope of the tangent line to the graph of f(x)=x2x+1f(x)=x^2/(x+1) at (2, 4/3)?