Matematyka
$f\left(x\right) =$ |
$\dfrac{\sqrt{x}}{\sqrt{x}+1}$
Note: Your input has been rewritten/simplified. |
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$\dfrac{\mathrm{d}\left(f\left(x\right)\right)}{\mathrm{d}x} =$ |
$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{\sqrt{x}}{\sqrt{x}+1}\right)}}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-4}{\left(\sqrt{x}+1\right){\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sqrt{x}\right)}}}}-\class{steps-node}{\cssId{steps-node-6}{\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sqrt{x}+1\right)}}{\cdot}\sqrt{x}}}}{\class{steps-node}{\cssId{steps-node-2}{{\left(\sqrt{x}+1\right)}^{2}}}}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-7}{\dfrac{1}{2{\cdot}\sqrt{x}}}}{\cdot}\left(\sqrt{x}+1\right)-\class{steps-node}{\cssId{steps-node-8}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sqrt{x}\right)}}{\cdot}\sqrt{x}}{{\left(\sqrt{x}+1\right)}^{2}}$ $=\dfrac{\dfrac{\sqrt{x}+1}{2{\cdot}\sqrt{x}}-\class{steps-node}{\cssId{steps-node-9}{\dfrac{1}{2{\cdot}\sqrt{x}}}}{\cdot}\sqrt{x}}{{\left(\sqrt{x}+1\right)}^{2}}$ $=\dfrac{\dfrac{\sqrt{x}+1}{2{\cdot}\sqrt{x}}-\dfrac{1}{2}}{{\left(\sqrt{x}+1\right)}^{2}}$ Wynik alternatywny: $=\dfrac{1}{2{\cdot}\left(\sqrt{x}+1\right){\cdot}\sqrt{x}}-\dfrac{1}{2{\cdot}{\left(\sqrt{x}+1\right)}^{2}}$ |