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Pochodna funkcji (1+1/x)^x

$f\left(x\right) =$	${\left(\dfrac{1}{x}+1\right)}^{x}$ Note: Your input has been rewritten/simplified.
$\dfrac{\mathrm{d}\left(f\left(x\right)\right)}{\mathrm{d}x} =$	$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({\left(\dfrac{1}{x}+1\right)}^{x}\right)}}$ $=\class{steps-node}{\cssId{steps-node-2}{{\left(\dfrac{1}{x}+1\right)}^{x}}}{\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\ln\left(\dfrac{1}{x}+1\right){\cdot}x\right)}}$ $=\left(\class{steps-node}{\cssId{steps-node-5}{\class{steps-node}{\cssId{steps-node-4}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\ln\left(\dfrac{1}{x}+1\right)\right)}}{\cdot}x}}+\class{steps-node}{\cssId{steps-node-7}{\ln\left(\dfrac{1}{x}+1\right){\cdot}\class{steps-node}{\cssId{steps-node-6}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x\right)}}}}\right){\cdot}{\left(\dfrac{1}{x}+1\right)}^{x}$ $={\left(\dfrac{1}{x}+1\right)}^{x}{\cdot}\left(\class{steps-node}{\cssId{steps-node-8}{\dfrac{1}{\dfrac{1}{x}+1}}}{\cdot}\class{steps-node}{\cssId{steps-node-9}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{1}{x}+1\right)}}{\cdot}x+\class{steps-node}{\cssId{steps-node-10}{1}}{\cdot}\ln\left(\dfrac{1}{x}+1\right)\right)$ $={\left(\dfrac{1}{x}+1\right)}^{x}{\cdot}\left(\dfrac{\class{steps-node}{\cssId{steps-node-11}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{1}{x}\right)}}{\cdot}x}{\dfrac{1}{x}+1}+\ln\left(\dfrac{1}{x}+1\right)\right)$ $={\left(\dfrac{1}{x}+1\right)}^{x}{\cdot}\left(\dfrac{\dfrac{\class{steps-node}{\cssId{steps-node-14}{-\class{steps-node}{\cssId{steps-node-13}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x\right)}}}}}{\class{steps-node}{\cssId{steps-node-12}{{x}^{2}}}}{\cdot}x}{\dfrac{1}{x}+1}+\ln\left(\dfrac{1}{x}+1\right)\right)$ $={\left(\dfrac{1}{x}+1\right)}^{x}{\cdot}\left(\ln\left(\dfrac{1}{x}+1\right)-\dfrac{\class{steps-node}{\cssId{steps-node-15}{1}}}{\left(\dfrac{1}{x}+1\right){\cdot}x}\right)$

$f\left(x\right) =$

${\left(\dfrac{1}{x}+1\right)}^{x}$

Note: Your input has been rewritten/simplified.

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

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

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

$=\left(\class{steps-node}{\cssId{steps-node-5}{\class{steps-node}{\cssId{steps-node-4}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\ln\left(\dfrac{1}{x}+1\right)\right)}}{\cdot}x}}+\class{steps-node}{\cssId{steps-node-7}{\ln\left(\dfrac{1}{x}+1\right){\cdot}\class{steps-node}{\cssId{steps-node-6}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x\right)}}}}\right){\cdot}{\left(\dfrac{1}{x}+1\right)}^{x}$

$={\left(\dfrac{1}{x}+1\right)}^{x}{\cdot}\left(\class{steps-node}{\cssId{steps-node-8}{\dfrac{1}{\dfrac{1}{x}+1}}}{\cdot}\class{steps-node}{\cssId{steps-node-9}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{1}{x}+1\right)}}{\cdot}x+\class{steps-node}{\cssId{steps-node-10}{1}}{\cdot}\ln\left(\dfrac{1}{x}+1\right)\right)$

$={\left(\dfrac{1}{x}+1\right)}^{x}{\cdot}\left(\dfrac{\class{steps-node}{\cssId{steps-node-11}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{1}{x}\right)}}{\cdot}x}{\dfrac{1}{x}+1}+\ln\left(\dfrac{1}{x}+1\right)\right)$

$={\left(\dfrac{1}{x}+1\right)}^{x}{\cdot}\left(\dfrac{\dfrac{\class{steps-node}{\cssId{steps-node-14}{-\class{steps-node}{\cssId{steps-node-13}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x\right)}}}}}{\class{steps-node}{\cssId{steps-node-12}{{x}^{2}}}}{\cdot}x}{\dfrac{1}{x}+1}+\ln\left(\dfrac{1}{x}+1\right)\right)$

$={\left(\dfrac{1}{x}+1\right)}^{x}{\cdot}\left(\ln\left(\dfrac{1}{x}+1\right)-\dfrac{\class{steps-node}{\cssId{steps-node-15}{1}}}{\left(\dfrac{1}{x}+1\right){\cdot}x}\right)$