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Pochodna funkcji [1+sin(2x)]/[1-sin(2x)]

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

$f\left(x\right) =$

$\dfrac{\sin\left(2x\right)+1}{1-\sin\left(2x\right)}$

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

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

$=\dfrac{\class{steps-node}{\cssId{steps-node-4}{\left(1-\sin\left(2x\right)\right){\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sin\left(2x\right)+1\right)}}}}-\class{steps-node}{\cssId{steps-node-6}{\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(1-\sin\left(2x\right)\right)}}{\cdot}\left(\sin\left(2x\right)+1\right)}}}{\class{steps-node}{\cssId{steps-node-2}{{\left(1-\sin\left(2x\right)\right)}^{2}}}}$

$=\dfrac{\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sin\left(2x\right)\right)}}{\cdot}\left(1-\sin\left(2x\right)\right)-\class{steps-node}{\cssId{steps-node-8}{-\class{steps-node}{\cssId{steps-node-9}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sin\left(2x\right)\right)}}}}{\cdot}\left(\sin\left(2x\right)+1\right)}{{\left(1-\sin\left(2x\right)\right)}^{2}}$

$=\dfrac{\class{steps-node}{\cssId{steps-node-12}{\cos\left(2x\right)}}{\cdot}\class{steps-node}{\cssId{steps-node-13}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(2x\right)}}{\cdot}\left(\sin\left(2x\right)+1\right)+\class{steps-node}{\cssId{steps-node-10}{\cos\left(2x\right)}}{\cdot}\class{steps-node}{\cssId{steps-node-11}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(2x\right)}}{\cdot}\left(1-\sin\left(2x\right)\right)}{{\left(1-\sin\left(2x\right)\right)}^{2}}$

$=\dfrac{\class{steps-node}{\cssId{steps-node-14}{2}}{\cdot}\cos\left(2x\right){\cdot}\left(\sin\left(2x\right)+1\right)+\class{steps-node}{\cssId{steps-node-15}{2}}{\cdot}\cos\left(2x\right){\cdot}\left(1-\sin\left(2x\right)\right)}{{\left(1-\sin\left(2x\right)\right)}^{2}}$

Wynik alternatywny:

$=\dfrac{2{\cdot}\cos\left(2x\right){\cdot}\left(\sin\left(2x\right)+1\right)}{{\left(1-\sin\left(2x\right)\right)}^{2}}+\dfrac{2{\cdot}\cos\left(2x\right)}{1-\sin\left(2x\right)}$