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Pochodna funkcji (sqrt(x))/(x+1)

$f\left(x\right) =$	$\dfrac{\sqrt{x}}{x+1}$
$\dfrac{\mathrm{d}\left(f\left(x\right)\right)}{\mathrm{d}x} =$	$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{\sqrt{x}}{x+1}\right)}}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-4}{\left(x+1\right){\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sqrt{x}\right)}}}}-\class{steps-node}{\cssId{steps-node-6}{\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x+1\right)}}{\cdot}\sqrt{x}}}}{\class{steps-node}{\cssId{steps-node-2}{{\left(x+1\right)}^{2}}}}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-7}{\dfrac{1}{2{\cdot}\sqrt{x}}}}{\cdot}\left(x+1\right)-\class{steps-node}{\cssId{steps-node-8}{1}}{\cdot}\sqrt{x}}{{\left(x+1\right)}^{2}}$ $=\dfrac{\dfrac{x+1}{2{\cdot}\sqrt{x}}-\sqrt{x}}{{\left(x+1\right)}^{2}}$ Wynik alternatywny: $=\dfrac{1}{2{\cdot}\sqrt{x}{\cdot}\left(x+1\right)}-\dfrac{\sqrt{x}}{{\left(x+1\right)}^{2}}$

$f\left(x\right) =$

$\dfrac{\sqrt{x}}{x+1}$

$\dfrac{\mathrm{d}\left(f\left(x\right)\right)}{\mathrm{d}x} =$

$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{\sqrt{x}}{x+1}\right)}}$

$=\dfrac{\class{steps-node}{\cssId{steps-node-4}{\left(x+1\right){\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sqrt{x}\right)}}}}-\class{steps-node}{\cssId{steps-node-6}{\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x+1\right)}}{\cdot}\sqrt{x}}}}{\class{steps-node}{\cssId{steps-node-2}{{\left(x+1\right)}^{2}}}}$

$=\dfrac{\class{steps-node}{\cssId{steps-node-7}{\dfrac{1}{2{\cdot}\sqrt{x}}}}{\cdot}\left(x+1\right)-\class{steps-node}{\cssId{steps-node-8}{1}}{\cdot}\sqrt{x}}{{\left(x+1\right)}^{2}}$

$=\dfrac{\dfrac{x+1}{2{\cdot}\sqrt{x}}-\sqrt{x}}{{\left(x+1\right)}^{2}}$

Wynik alternatywny:

$=\dfrac{1}{2{\cdot}\sqrt{x}{\cdot}\left(x+1\right)}-\dfrac{\sqrt{x}}{{\left(x+1\right)}^{2}}$