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Pochodna funkcji arccos(sqrt(-x+1))

$f\left(x\right) =$	$\arccos\left(\sqrt{1-x}\right)$ Note: Your input has been rewritten/simplified.
$\dfrac{\mathrm{d}\left(f\left(x\right)\right)}{\mathrm{d}x} =$	$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\arccos\left(\sqrt{1-x}\right)\right)}}$ $=\class{steps-node}{\cssId{steps-node-2}{\dfrac{-1}{\sqrt{1-1-x}}}}{\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sqrt{1-x}\right)}}$ $=\dfrac{-\left(\class{steps-node}{\cssId{steps-node-4}{\dfrac{1}{2{\cdot}\sqrt{1-x}}}}{\cdot}\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(1-x\right)}}\right)}{\sqrt{x}}$ $=\dfrac{-\class{steps-node}{\cssId{steps-node-6}{\left(-1\right)}}}{2{\cdot}\sqrt{1-x}{\cdot}\sqrt{x}}$ $=\dfrac{1}{2{\cdot}\sqrt{1-x}{\cdot}\sqrt{x}}$

$f\left(x\right) =$

$\arccos\left(\sqrt{1-x}\right)$

Note: Your input has been rewritten/simplified.

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

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

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

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

$=\dfrac{-\class{steps-node}{\cssId{steps-node-6}{\left(-1\right)}}}{2{\cdot}\sqrt{1-x}{\cdot}\sqrt{x}}$

$=\dfrac{1}{2{\cdot}\sqrt{1-x}{\cdot}\sqrt{x}}$