Define the function $f$ 

$f\left(x\right)\=\frac{x^2\+1}{x\+2}$$\stackrel{\text{assign as function}}{\to }$$f$

Method 1

$f\prime \left(c\right)$ = $\frac{2c}{c\+2}-\frac{c^2\+1}{{\left(c\+2\right)}^2}$$\stackrel{\text{simplify}}{\=}$$\frac{c^2\+4c-1}{{\left(c\+2\right)}^2}$$$

Method 2

$f\left(x\right)$

$\frac{x^2\+1}{x\+2}$

$\stackrel{\text{differentiate w.r.t. x}}{\to }$

$\frac{2x}{x\+2}-\frac{x^2\+1}{{\left(x\+2\right)}^2}$

$\stackrel{\text{simplify}}{\=}$

$\frac{x^2\+4x-1}{{\left(x\+2\right)}^2}$

$\stackrel{\text{evaluate at point}}{\to }$

$\frac{c^2\+4c-1}{{\left(c\+2\right)}^2}$

Method 3

Context Panel: Simplify≻Simplify

$\frac{\ⅆ}{\ⅆ x} f\left(x\right)$

$\frac{2x}{x\+2}-\frac{x^2\+1}{{\left(x\+2\right)}^2}$

$\stackrel{\text{simplify}}{\=}$

$\frac{x^2\+4x-1}{{\left(x\+2\right)}^2}$

$\stackrel{\text{evaluate at point}}{\to }$

$\frac{c^2\+4c-1}{{\left(c\+2\right)}^2}$

Stepwise solutions

Loading Student:-Calculus1

$\mathrm{NewtonQuotient}\left(f\left(x\right)\,x\=c\,h\=h\right)$

$\frac{c^2\+ch\+4c\+2h-1}{\left(c\+h\+2\right)\left(c\+2\right)}$

$\stackrel{\text{evaluate at point}}{\to }$

$\frac{c^2\+4c-1}{{\left(c\+2\right)}^2}$

Application of Definition 2.2.1

$\underset{h\to 0}{lim}\frac{f\left(c\+h\right)-f\left(c\right)}{h}$ = $\frac{c^2\+4c-1}{{\left(c\+2\right)}^2}$$$

The difference quotient and its simplification

$\frac{f\left(c\+h\right)-f\left(c\right)}{h}$

$\frac{\frac{{\left(c\+h\right)}^2\+1}{c\+h\+2}-\frac{c^2\+1}{c\+2}}{h}$

$\stackrel{\text{simplify}}{\=}$

$\frac{c^2\+ch\+4c\+2h-1}{\left(c\+h\+2\right)\left(c\+2\right)}$

$\stackrel{\text{evaluate at point}}{\to }$

$\frac{c^2\+4c-1}{{\left(c\+2\right)}^2}$

$\frac{\frac{{\left(c\+h\right)}^2\+1}{c\+h\+2}-\frac{c^2\+1}{c\+2}}{h}$

$\=\frac{\frac{\left({\left(c\+h\right)}^2\+1\right)\cdot \left(c\+2\right)-\left(c^2\+1\right)\cdot \left(c\+h\+2\right)}{\left(c\+h\+2\right)\cdot \left(c\+2\right)}}{h}$

$\=\frac{A-B}{h \left(c\+h\+2\right)\cdot \left(c\+2\right)}$, where

$A\=c^3\+2c^2\+2c^{2}h\+4ch\+h^{2}c\+2h^2\+c\+2$ 

$B\=c^3\+c^{2}h\+2c^2\+c\+h\+2$

$\=\frac{h \left(c^2\+4c\+ch\+2h-1\right)}{h \left(c\+h\+2\right)\cdot \left(c\+2\right)}$

$\=\frac{c^2\+4c\+ch\+2h-1}{\left(c\+h\+2\right)\cdot \left(c\+2\right)}$