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Pochodna funkcji (lnx)/(x^3+1)

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

$f\left(x\right) =$

$\dfrac{\ln\left(x\right)}{{x}^{3}+1}$

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

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

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

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

$=\dfrac{\dfrac{{x}^{3}+1}{x}-\class{steps-node}{\cssId{steps-node-9}{3}}\class{steps-node}{\cssId{steps-node-10}{{x}^{2}}}{\cdot}\ln\left(x\right)}{{\left({x}^{3}+1\right)}^{2}}$

Wynik alternatywny:

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