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Pochodna funkcji arccos(sqrt((1-xx)/(1+xx)))

$f\left(x\right) =$	$\arccos\left(\dfrac{\sqrt{1-{x}^{2}}}{\sqrt{{x}^{2}+1}}\right)$ Note: Your input has been rewritten/simplified.
$\dfrac{\mathrm{d}\left(f\left(x\right)\right)}{\mathrm{d}x} =$	$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\arccos\left(\dfrac{\sqrt{1-{x}^{2}}}{\sqrt{{x}^{2}+1}}\right)\right)}}$ $=\class{steps-node}{\cssId{steps-node-2}{\dfrac{-1}{\sqrt{1-{\left(\dfrac{\sqrt{1-{x}^{2}}}{\sqrt{{x}^{2}+1}}\right)}^{2}}}}}{\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{\sqrt{1-{x}^{2}}}{\sqrt{{x}^{2}+1}}\right)}}$ $=\dfrac{-\dfrac{\class{steps-node}{\cssId{steps-node-6}{\sqrt{{x}^{2}+1}{\cdot}\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sqrt{1-{x}^{2}}\right)}}}}-\class{steps-node}{\cssId{steps-node-8}{\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sqrt{{x}^{2}+1}\right)}}{\cdot}\sqrt{1-{x}^{2}}}}}{\class{steps-node}{\cssId{steps-node-4}{{x}^{2}+1}}}}{\sqrt{1-\dfrac{1-{x}^{2}}{{x}^{2}+1}}}$ $=\dfrac{-\left(\class{steps-node}{\cssId{steps-node-9}{\dfrac{1}{2{\cdot}\sqrt{1-{x}^{2}}}}}{\cdot}\class{steps-node}{\cssId{steps-node-10}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(1-{x}^{2}\right)}}{\cdot}\sqrt{{x}^{2}+1}-\class{steps-node}{\cssId{steps-node-11}{\dfrac{1}{2{\cdot}\sqrt{{x}^{2}+1}}}}{\cdot}\class{steps-node}{\cssId{steps-node-12}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}+1\right)}}{\cdot}\sqrt{1-{x}^{2}}\right)}{\left({x}^{2}+1\right){\cdot}\sqrt{1-\dfrac{1-{x}^{2}}{{x}^{2}+1}}}$ $=\dfrac{\dfrac{\class{steps-node}{\cssId{steps-node-15}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}\right)}}{\cdot}\sqrt{1-{x}^{2}}}{2{\cdot}\sqrt{{x}^{2}+1}}-\dfrac{\class{steps-node}{\cssId{steps-node-13}{-\class{steps-node}{\cssId{steps-node-14}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}\right)}}}}{\cdot}\sqrt{{x}^{2}+1}}{2{\cdot}\sqrt{1-{x}^{2}}}}{\left({x}^{2}+1\right){\cdot}\sqrt{1-\dfrac{1-{x}^{2}}{{x}^{2}+1}}}$ $=\dfrac{\dfrac{\class{steps-node}{\cssId{steps-node-18}{2}}\class{steps-node}{\cssId{steps-node-19}{x}}{\cdot}\sqrt{{x}^{2}+1}}{2{\cdot}\sqrt{1-{x}^{2}}}+\dfrac{\class{steps-node}{\cssId{steps-node-16}{2}}\class{steps-node}{\cssId{steps-node-17}{x}}{\cdot}\sqrt{1-{x}^{2}}}{2{\cdot}\sqrt{{x}^{2}+1}}}{\left({x}^{2}+1\right){\cdot}\sqrt{1-\dfrac{1-{x}^{2}}{{x}^{2}+1}}}$ $=\dfrac{\dfrac{x{\cdot}\sqrt{{x}^{2}+1}}{\sqrt{1-{x}^{2}}}+\dfrac{x{\cdot}\sqrt{1-{x}^{2}}}{\sqrt{{x}^{2}+1}}}{\left({x}^{2}+1\right){\cdot}\sqrt{1-\dfrac{1-{x}^{2}}{{x}^{2}+1}}}$ Uproszczony wynik: $=\dfrac{-\left(\dfrac{-x}{\sqrt{1-{x}^{2}}{\cdot}\sqrt{{x}^{2}+1}}-\dfrac{x{\cdot}\sqrt{1-{x}^{2}}}{{\left({x}^{2}+1\right)}^{\frac{3}{2}}}\right)}{\sqrt{1-\dfrac{1-{x}^{2}}{{x}^{2}+1}}}$

$f\left(x\right) =$

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

Note: Your input has been rewritten/simplified.

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

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

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

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

$=\dfrac{-\left(\class{steps-node}{\cssId{steps-node-9}{\dfrac{1}{2{\cdot}\sqrt{1-{x}^{2}}}}}{\cdot}\class{steps-node}{\cssId{steps-node-10}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(1-{x}^{2}\right)}}{\cdot}\sqrt{{x}^{2}+1}-\class{steps-node}{\cssId{steps-node-11}{\dfrac{1}{2{\cdot}\sqrt{{x}^{2}+1}}}}{\cdot}\class{steps-node}{\cssId{steps-node-12}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}+1\right)}}{\cdot}\sqrt{1-{x}^{2}}\right)}{\left({x}^{2}+1\right){\cdot}\sqrt{1-\dfrac{1-{x}^{2}}{{x}^{2}+1}}}$

$=\dfrac{\dfrac{\class{steps-node}{\cssId{steps-node-15}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}\right)}}{\cdot}\sqrt{1-{x}^{2}}}{2{\cdot}\sqrt{{x}^{2}+1}}-\dfrac{\class{steps-node}{\cssId{steps-node-13}{-\class{steps-node}{\cssId{steps-node-14}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}\right)}}}}{\cdot}\sqrt{{x}^{2}+1}}{2{\cdot}\sqrt{1-{x}^{2}}}}{\left({x}^{2}+1\right){\cdot}\sqrt{1-\dfrac{1-{x}^{2}}{{x}^{2}+1}}}$

$=\dfrac{\dfrac{\class{steps-node}{\cssId{steps-node-18}{2}}\class{steps-node}{\cssId{steps-node-19}{x}}{\cdot}\sqrt{{x}^{2}+1}}{2{\cdot}\sqrt{1-{x}^{2}}}+\dfrac{\class{steps-node}{\cssId{steps-node-16}{2}}\class{steps-node}{\cssId{steps-node-17}{x}}{\cdot}\sqrt{1-{x}^{2}}}{2{\cdot}\sqrt{{x}^{2}+1}}}{\left({x}^{2}+1\right){\cdot}\sqrt{1-\dfrac{1-{x}^{2}}{{x}^{2}+1}}}$

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

Uproszczony wynik:

$=\dfrac{-\left(\dfrac{-x}{\sqrt{1-{x}^{2}}{\cdot}\sqrt{{x}^{2}+1}}-\dfrac{x{\cdot}\sqrt{1-{x}^{2}}}{{\left({x}^{2}+1\right)}^{\frac{3}{2}}}\right)}{\sqrt{1-\dfrac{1-{x}^{2}}{{x}^{2}+1}}}$