Matematyka
Pochodna funkcji ln(x+sqrt(1+x^2))
| $f\left(x\right) =$ |
$\ln\left(\sqrt{{x}^{2}+1}+x\right)$
Note: Your input has been rewritten/simplified. |
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| $\dfrac{\mathrm{d}\left(f\left(x\right)\right)}{\mathrm{d}x} =$ |
$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\ln\left(\sqrt{{x}^{2}+1}+x\right)\right)}}$ $=\class{steps-node}{\cssId{steps-node-2}{\dfrac{1}{\sqrt{{x}^{2}+1}+x}}}{\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sqrt{{x}^{2}+1}+x\right)}}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-4}{\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sqrt{{x}^{2}+1}\right)}}+1}}}{\sqrt{{x}^{2}+1}+x}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-6}{\dfrac{1}{2{\cdot}\sqrt{{x}^{2}+1}}}}{\cdot}\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}+1\right)}}+1}{\sqrt{{x}^{2}+1}+x}$ $=\dfrac{\dfrac{\class{steps-node}{\cssId{steps-node-8}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}\right)}}}{2{\cdot}\sqrt{{x}^{2}+1}}+1}{\sqrt{{x}^{2}+1}+x}$ $=\dfrac{\dfrac{\class{steps-node}{\cssId{steps-node-9}{2}}\class{steps-node}{\cssId{steps-node-10}{x}}}{2{\cdot}\sqrt{{x}^{2}+1}}+1}{\sqrt{{x}^{2}+1}+x}$ $=\dfrac{\dfrac{x}{\sqrt{{x}^{2}+1}}+1}{\sqrt{{x}^{2}+1}+x}$ |
