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Pochodna funkcji cos(2x+5)

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

$f\left(x\right) =$

$\cos\left(2x+5\right)$

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

$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathbf{d}}{\mathbf{d}\boldsymbol{x}}\kern-.25em\left(\cos\left(2x+5\right)\right)}}$

$=\class{steps-node}{\cssId{steps-node-2}{-\sin\left(2x+5\right)}}{\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathbf{d}}{\mathbf{d}\boldsymbol{x}}\kern-.25em\left(2x+5\right)}}$

$=-\class{steps-node}{\cssId{steps-node-4}{2}}{\cdot}\sin\left(2x+5\right)$