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Pochodna funkcji (e^x/x^2)*sin(2x)

$f\left(x\right) =$	$\dfrac{{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{2}}$ Note: Your input has been rewritten/simplified.
$\dfrac{\mathrm{d}\left(f\left(x\right)\right)}{\mathrm{d}x} =$	$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{2}}\right)}}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-4}{{x}^{2}{\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)\right)}}}}-\class{steps-node}{\cssId{steps-node-6}{\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}\right)}}{\cdot}{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}}}{\class{steps-node}{\cssId{steps-node-2}{{\left({x}^{2}\right)}^{2}}}}$ $=\dfrac{\left(\class{steps-node}{\cssId{steps-node-8}{\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({\mathrm{e}}^{x}\right)}}{\cdot}\sin\left(2x\right)}}+\class{steps-node}{\cssId{steps-node-10}{{\mathrm{e}}^{x}{\cdot}\class{steps-node}{\cssId{steps-node-9}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sin\left(2x\right)\right)}}}}\right){\cdot}{x}^{2}-\class{steps-node}{\cssId{steps-node-11}{2}}\class{steps-node}{\cssId{steps-node-12}{x}}{\cdot}{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{4}}$ $=\dfrac{{x}^{2}{\cdot}\left(\class{steps-node}{\cssId{steps-node-13}{{\mathrm{e}}^{x}}}{\cdot}\sin\left(2x\right)+\class{steps-node}{\cssId{steps-node-14}{\cos\left(2x\right)}}{\cdot}\class{steps-node}{\cssId{steps-node-15}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(2x\right)}}{\cdot}{\mathrm{e}}^{x}\right)-2x{\cdot}{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{4}}$ $=\dfrac{{x}^{2}{\cdot}\left({\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)+\class{steps-node}{\cssId{steps-node-16}{2}}{\mathrm{e}}^{x}{\cdot}\cos\left(2x\right)\right)-2x{\cdot}{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{4}}$ Wynik alternatywny: $=\dfrac{{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{2}}-\dfrac{2{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{3}}+\dfrac{2{\mathrm{e}}^{x}{\cdot}\cos\left(2x\right)}{{x}^{2}}$

$f\left(x\right) =$

$\dfrac{{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{2}}$

Note: Your input has been rewritten/simplified.

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

$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{2}}\right)}}$

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

$=\dfrac{\left(\class{steps-node}{\cssId{steps-node-8}{\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({\mathrm{e}}^{x}\right)}}{\cdot}\sin\left(2x\right)}}+\class{steps-node}{\cssId{steps-node-10}{{\mathrm{e}}^{x}{\cdot}\class{steps-node}{\cssId{steps-node-9}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sin\left(2x\right)\right)}}}}\right){\cdot}{x}^{2}-\class{steps-node}{\cssId{steps-node-11}{2}}\class{steps-node}{\cssId{steps-node-12}{x}}{\cdot}{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{4}}$

$=\dfrac{{x}^{2}{\cdot}\left(\class{steps-node}{\cssId{steps-node-13}{{\mathrm{e}}^{x}}}{\cdot}\sin\left(2x\right)+\class{steps-node}{\cssId{steps-node-14}{\cos\left(2x\right)}}{\cdot}\class{steps-node}{\cssId{steps-node-15}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(2x\right)}}{\cdot}{\mathrm{e}}^{x}\right)-2x{\cdot}{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{4}}$

$=\dfrac{{x}^{2}{\cdot}\left({\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)+\class{steps-node}{\cssId{steps-node-16}{2}}{\mathrm{e}}^{x}{\cdot}\cos\left(2x\right)\right)-2x{\cdot}{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{4}}$

Wynik alternatywny:

$=\dfrac{{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{2}}-\dfrac{2{\mathrm{e}}^{x}{\cdot}\sin\left(2x\right)}{{x}^{3}}+\dfrac{2{\mathrm{e}}^{x}{\cdot}\cos\left(2x\right)}{{x}^{2}}$