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Pochodna funkcji arctg(1-2cosx)/(pi-3x)

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

$f\left(a, c, g, r, t, x\right) =$

$\dfrac{acgrt{\cdot}\left(1-2{\cdot}\cos\left(x\right)\right)}{{\pi}-3x}$

Note: Your input has been rewritten/simplified.

$\dfrac{\mathrm{d}\left(f\left(a, c, g, r, t, x\right)\right)}{\mathrm{d}x} =$

$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{acgrt{\cdot}\left(1-2{\cdot}\cos\left(x\right)\right)}{{\pi}-3x}\right)}}$

$=\class{steps-node}{\cssId{steps-node-2}{acgrt{\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{1-2{\cdot}\cos\left(x\right)}{{\pi}-3x}\right)}}}}$

$=acgrt{\cdot}\dfrac{\class{steps-node}{\cssId{steps-node-6}{\left({\pi}-3x\right){\cdot}\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(1-2{\cdot}\cos\left(x\right)\right)}}}}-\class{steps-node}{\cssId{steps-node-8}{\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({\pi}-3x\right)}}{\cdot}\left(1-2{\cdot}\cos\left(x\right)\right)}}}{\class{steps-node}{\cssId{steps-node-4}{{\left({\pi}-3x\right)}^{2}}}}$

$=\dfrac{acgrt{\cdot}\left(\class{steps-node}{\cssId{steps-node-9}{-2{\cdot}\class{steps-node}{\cssId{steps-node-10}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\cos\left(x\right)\right)}}}}{\cdot}\left({\pi}-3x\right)-\class{steps-node}{\cssId{steps-node-11}{-3}}{\cdot}\left(1-2{\cdot}\cos\left(x\right)\right)\right)}{{\left({\pi}-3x\right)}^{2}}$

$=\dfrac{acgrt{\cdot}\left(3{\cdot}\left(1-2{\cdot}\cos\left(x\right)\right)-2{\cdot}\class{steps-node}{\cssId{steps-node-12}{\left(-\sin\left(x\right)\right)}}{\cdot}\left({\pi}-3x\right)\right)}{{\left({\pi}-3x\right)}^{2}}$

$=\dfrac{acgrt{\cdot}\left(2{\cdot}\left({\pi}-3x\right){\cdot}\sin\left(x\right)+3{\cdot}\left(1-2{\cdot}\cos\left(x\right)\right)\right)}{{\left({\pi}-3x\right)}^{2}}$

Wynik alternatywny:

$=\dfrac{2acgrt{\cdot}\sin\left(x\right)}{{\pi}-3x}+\dfrac{3acgrt{\cdot}\left(1-2{\cdot}\cos\left(x\right)\right)}{{\left({\pi}-3x\right)}^{2}}$