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Pochodna funkcji arcsin(2x(sqrt(1-x)))

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

$f\left(x\right) =$

$\arcsin\left(2{\cdot}\sqrt{1-x}{\cdot}x\right)$

Note: Your input has been rewritten/simplified.

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

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

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

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

$=\dfrac{2{\cdot}\left(\class{steps-node}{\cssId{steps-node-7}{\class{steps-node}{\cssId{steps-node-6}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sqrt{1-x}\right)}}{\cdot}x}}+\class{steps-node}{\cssId{steps-node-9}{\sqrt{1-x}{\cdot}\class{steps-node}{\cssId{steps-node-8}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x\right)}}}}\right)}{\sqrt{1-4{\cdot}\left(1-x\right){\cdot}{x}^{2}}}$

$=\dfrac{2{\cdot}\left(\class{steps-node}{\cssId{steps-node-10}{\dfrac{1}{2{\cdot}\sqrt{1-x}}}}{\cdot}\class{steps-node}{\cssId{steps-node-11}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(1-x\right)}}{\cdot}x+\class{steps-node}{\cssId{steps-node-12}{1}}{\cdot}\sqrt{1-x}\right)}{\sqrt{1-4{\cdot}\left(1-x\right){\cdot}{x}^{2}}}$

$=\dfrac{2{\cdot}\left(\dfrac{\class{steps-node}{\cssId{steps-node-13}{-1}}x}{2{\cdot}\sqrt{1-x}}+\sqrt{1-x}\right)}{\sqrt{1-4{\cdot}\left(1-x\right){\cdot}{x}^{2}}}$

$=\dfrac{2{\cdot}\left(\sqrt{1-x}-\dfrac{x}{2{\cdot}\sqrt{1-x}}\right)}{\sqrt{1-4{\cdot}\left(1-x\right){\cdot}{x}^{2}}}$

Uproszczony wynik:

$=\dfrac{2{\cdot}\sqrt{1-x}-\dfrac{x}{\sqrt{1-x}}}{\sqrt{1-4{\cdot}\left(1-x\right){\cdot}{x}^{2}}}$