Pochodna funkcji (e^x+1)/(e^x+x)

$f\left(x\right) =$ $\dfrac{{\mathrm{e}}^{x}+1}{{\mathrm{e}}^{x}+x}$
$\dfrac{\mathrm{d}\left(f\left(x\right)\right)}{\mathrm{d}x} =$

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

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

$=\dfrac{\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({\mathrm{e}}^{x}\right)}}{\cdot}\left({\mathrm{e}}^{x}+x\right)-\class{steps-node}{\cssId{steps-node-8}{\left(\class{steps-node}{\cssId{steps-node-9}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({\mathrm{e}}^{x}\right)}}+1\right)}}{\cdot}\left({\mathrm{e}}^{x}+1\right)}{{\left({\mathrm{e}}^{x}+x\right)}^{2}}$

$=\dfrac{\class{steps-node}{\cssId{steps-node-10}{{\mathrm{e}}^{x}}}{\cdot}\left({\mathrm{e}}^{x}+x\right)-\left(\class{steps-node}{\cssId{steps-node-11}{{\mathrm{e}}^{x}}}+1\right){\cdot}\left({\mathrm{e}}^{x}+1\right)}{{\left({\mathrm{e}}^{x}+x\right)}^{2}}$

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

Uproszczony wynik:

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

Podziel się rozwiązaniem:

Wybrane przykłady