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

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

$f\left(x\right) =$

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

Note: Your input has been rewritten/simplified.

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

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

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

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

$=\dfrac{\dfrac{\class{steps-node}{\cssId{steps-node-8}{\left(x+1\right){\cdot}\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(1-x\right)}}}}-\class{steps-node}{\cssId{steps-node-10}{\class{steps-node}{\cssId{steps-node-9}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x+1\right)}}{\cdot}\left(1-x\right)}}}{\class{steps-node}{\cssId{steps-node-6}{{\left(x+1\right)}^{2}}}}}{2{\cdot}\sqrt{\dfrac{1-x}{x+1}}{\cdot}\left(\dfrac{1-x}{x+1}+1\right)}$

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

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

Wynik alternatywny:

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