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Pochodna funkcji arctan(x-(sqrt(x*x+1)))

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

$f\left(x\right) =$

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

Note: Your input has been rewritten/simplified.

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

$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(-\arctan\left(\sqrt{{x}^{2}+1}-x\right)\right)}}$

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

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

$=\dfrac{-\class{steps-node}{\cssId{steps-node-6}{\left(\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sqrt{{x}^{2}+1}\right)}}-1\right)}}}{{\left(\sqrt{{x}^{2}+1}-x\right)}^{2}+1}$

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

$=\dfrac{1-\dfrac{\class{steps-node}{\cssId{steps-node-10}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}\right)}}}{2{\cdot}\sqrt{{x}^{2}+1}}}{{\left(\sqrt{{x}^{2}+1}-x\right)}^{2}+1}$

$=\dfrac{1-\dfrac{\class{steps-node}{\cssId{steps-node-11}{2}}\class{steps-node}{\cssId{steps-node-12}{x}}}{2{\cdot}\sqrt{{x}^{2}+1}}}{{\left(\sqrt{{x}^{2}+1}-x\right)}^{2}+1}$

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

Wynik alternatywny:

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