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

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

$f\left(x\right) =$

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

Note: Your input has been rewritten/simplified.

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

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

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

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

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

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

$=\dfrac{2{\cdot}\left(\sqrt{1-{x}^{2}}+\dfrac{\class{steps-node}{\cssId{steps-node-13}{-\class{steps-node}{\cssId{steps-node-14}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}\right)}}}}{\cdot}x}{2{\cdot}\sqrt{1-{x}^{2}}}\right)}{\sqrt{1-4{x}^{2}{\cdot}\left(1-{x}^{2}\right)}}$

$=\dfrac{2{\cdot}\left(\sqrt{1-{x}^{2}}-\dfrac{\class{steps-node}{\cssId{steps-node-15}{2}}\class{steps-node}{\cssId{steps-node-16}{x}}{\cdot}x}{2{\cdot}\sqrt{1-{x}^{2}}}\right)}{\sqrt{1-4{x}^{2}{\cdot}\left(1-{x}^{2}\right)}}$

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

Wynik alternatywny:

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