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Pochodna funkcji (1-x^3)/(2-x)

$f\left(x\right) =$	$\dfrac{1-{x}^{3}}{2-x}$
$\dfrac{\mathrm{d}\left(f\left(x\right)\right)}{\mathrm{d}x} =$	$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{1-{x}^{3}}{2-x}\right)}}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-4}{\left(2-x\right){\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(1-{x}^{3}\right)}}}}-\class{steps-node}{\cssId{steps-node-6}{\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(2-x\right)}}{\cdot}\left(1-{x}^{3}\right)}}}{\class{steps-node}{\cssId{steps-node-2}{{\left(2-x\right)}^{2}}}}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-7}{-\class{steps-node}{\cssId{steps-node-8}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{3}\right)}}}}{\cdot}\left(2-x\right)-\class{steps-node}{\cssId{steps-node-9}{-1}}{\cdot}\left(1-{x}^{3}\right)}{{\left(2-x\right)}^{2}}$ $=\dfrac{-{x}^{3}-\class{steps-node}{\cssId{steps-node-10}{3}}\class{steps-node}{\cssId{steps-node-11}{{x}^{2}}}{\cdot}\left(2-x\right)+1}{{\left(2-x\right)}^{2}}$ $=\dfrac{-{x}^{3}-3{\cdot}\left(2-x\right){\cdot}{x}^{2}+1}{{\left(2-x\right)}^{2}}$ Uproszczony wynik: $=\dfrac{1-{x}^{3}}{{\left(2-x\right)}^{2}}-\dfrac{3{x}^{2}}{2-x}$

$f\left(x\right) =$

$\dfrac{1-{x}^{3}}{2-x}$

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

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

$=\dfrac{\class{steps-node}{\cssId{steps-node-4}{\left(2-x\right){\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(1-{x}^{3}\right)}}}}-\class{steps-node}{\cssId{steps-node-6}{\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(2-x\right)}}{\cdot}\left(1-{x}^{3}\right)}}}{\class{steps-node}{\cssId{steps-node-2}{{\left(2-x\right)}^{2}}}}$

$=\dfrac{\class{steps-node}{\cssId{steps-node-7}{-\class{steps-node}{\cssId{steps-node-8}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{3}\right)}}}}{\cdot}\left(2-x\right)-\class{steps-node}{\cssId{steps-node-9}{-1}}{\cdot}\left(1-{x}^{3}\right)}{{\left(2-x\right)}^{2}}$

$=\dfrac{-{x}^{3}-\class{steps-node}{\cssId{steps-node-10}{3}}\class{steps-node}{\cssId{steps-node-11}{{x}^{2}}}{\cdot}\left(2-x\right)+1}{{\left(2-x\right)}^{2}}$

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

Uproszczony wynik:

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