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