Matematyka
$f\left(x\right) =$ | $\dfrac{{x}^{3}-6}{{x}^{2}-5}$ |
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$\dfrac{\mathrm{d}\left(f\left(x\right)\right)}{\mathrm{d}x} =$ |
$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{{x}^{3}-6}{{x}^{2}-5}\right)}}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-4}{\left({x}^{2}-5\right){\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{3}-6\right)}}}}-\class{steps-node}{\cssId{steps-node-6}{\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}-5\right)}}{\cdot}\left({x}^{3}-6\right)}}}{\class{steps-node}{\cssId{steps-node-2}{{\left({x}^{2}-5\right)}^{2}}}}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{3}\right)}}{\cdot}\left({x}^{2}-5\right)-\class{steps-node}{\cssId{steps-node-8}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}\right)}}{\cdot}\left({x}^{3}-6\right)}{{\left({x}^{2}-5\right)}^{2}}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-9}{3}}\class{steps-node}{\cssId{steps-node-10}{{x}^{2}}}{\cdot}\left({x}^{2}-5\right)-\class{steps-node}{\cssId{steps-node-11}{2}}\class{steps-node}{\cssId{steps-node-12}{x}}{\cdot}\left({x}^{3}-6\right)}{{\left({x}^{2}-5\right)}^{2}}$ Wynik alternatywny: $=\dfrac{3{x}^{2}}{{x}^{2}-5}-\dfrac{2x{\cdot}\left({x}^{3}-6\right)}{{\left({x}^{2}-5\right)}^{2}}$ |