gdzie jesteś: Start / Pochodne / Pochodne funkcji trygonometrycznych / Pochodna funkcji 3cos(4x)+2sin(5x)-arccos(3x)+e^(x^2-2x+1)-(1/3)+ln(1-x^3)

Pochodna funkcji 3cos(4x)+2sin(5x)-arccos(3x)+e^(x^2-2x+1)-(1/3)+ln(1-x^3)

$f\left(x\right) =$	$\ln\left(1-{x}^{3}\right)+2{\cdot}\sin\left(5x\right)+3{\cdot}\cos\left(4x\right)-\arccos\left(3x\right)+{\mathrm{e}}^{{x}^{2}-2x+1}-\dfrac{1}{3}$ Note: Your input has been rewritten/simplified.
$\dfrac{\mathrm{d}\left(f\left(x\right)\right)}{\mathrm{d}x} =$	$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\ln\left(1-{x}^{3}\right)+2{\cdot}\sin\left(5x\right)+3{\cdot}\cos\left(4x\right)-\arccos\left(3x\right)+{\mathrm{e}}^{{x}^{2}-2x+1}-\dfrac{1}{3}\right)}}$ $=\class{steps-node}{\cssId{steps-node-2}{\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\ln\left(1-{x}^{3}\right)\right)}}+2{\cdot}\class{steps-node}{\cssId{steps-node-4}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sin\left(5x\right)\right)}}+3{\cdot}\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\cos\left(4x\right)\right)}}-\class{steps-node}{\cssId{steps-node-6}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\arccos\left(3x\right)\right)}}+\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({\mathrm{e}}^{{x}^{2}-2x+1}\right)}}}}$ $=\class{steps-node}{\cssId{steps-node-16}{{\mathrm{e}}^{{x}^{2}-2x+1}}}{\cdot}\class{steps-node}{\cssId{steps-node-17}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}-2x+1\right)}}-\class{steps-node}{\cssId{steps-node-14}{\dfrac{-1}{\sqrt{1-{\left(3x\right)}^{2}}}}}{\cdot}\class{steps-node}{\cssId{steps-node-15}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(3x\right)}}+3{\cdot}\class{steps-node}{\cssId{steps-node-12}{-\sin\left(4x\right)}}{\cdot}\class{steps-node}{\cssId{steps-node-13}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(4x\right)}}+2{\cdot}\class{steps-node}{\cssId{steps-node-10}{\cos\left(5x\right)}}{\cdot}\class{steps-node}{\cssId{steps-node-11}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(5x\right)}}+\class{steps-node}{\cssId{steps-node-8}{\dfrac{1}{1-{x}^{3}}}}{\cdot}\class{steps-node}{\cssId{steps-node-9}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(1-{x}^{3}\right)}}$ $=2{\cdot}\class{steps-node}{\cssId{steps-node-22}{5}}{\cdot}\cos\left(5x\right)-3{\cdot}\class{steps-node}{\cssId{steps-node-21}{4}}{\cdot}\sin\left(4x\right)+\class{steps-node}{\cssId{steps-node-18}{\left(\class{steps-node}{\cssId{steps-node-19}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}\right)}}-2\right)}}{\cdot}{\mathrm{e}}^{{x}^{2}-2x+1}+\dfrac{\class{steps-node}{\cssId{steps-node-23}{-\class{steps-node}{\cssId{steps-node-24}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{3}\right)}}}}}{1-{x}^{3}}+\dfrac{\class{steps-node}{\cssId{steps-node-20}{3}}}{\sqrt{1-9{x}^{2}}}$ $=10{\cdot}\cos\left(5x\right)-12{\cdot}\sin\left(4x\right)+\left(\class{steps-node}{\cssId{steps-node-25}{2}}\class{steps-node}{\cssId{steps-node-26}{x}}-2\right){\cdot}{\mathrm{e}}^{{x}^{2}-2x+1}-\dfrac{\class{steps-node}{\cssId{steps-node-27}{3}}\class{steps-node}{\cssId{steps-node-28}{{x}^{2}}}}{1-{x}^{3}}+\dfrac{3}{\sqrt{1-9{x}^{2}}}$

$f\left(x\right) =$

$\ln\left(1-{x}^{3}\right)+2{\cdot}\sin\left(5x\right)+3{\cdot}\cos\left(4x\right)-\arccos\left(3x\right)+{\mathrm{e}}^{{x}^{2}-2x+1}-\dfrac{1}{3}$

Note: Your input has been rewritten/simplified.

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

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

$=\class{steps-node}{\cssId{steps-node-2}{\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\ln\left(1-{x}^{3}\right)\right)}}+2{\cdot}\class{steps-node}{\cssId{steps-node-4}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sin\left(5x\right)\right)}}+3{\cdot}\class{steps-node}{\cssId{steps-node-5}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\cos\left(4x\right)\right)}}-\class{steps-node}{\cssId{steps-node-6}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\arccos\left(3x\right)\right)}}+\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({\mathrm{e}}^{{x}^{2}-2x+1}\right)}}}}$

$=\class{steps-node}{\cssId{steps-node-16}{{\mathrm{e}}^{{x}^{2}-2x+1}}}{\cdot}\class{steps-node}{\cssId{steps-node-17}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}-2x+1\right)}}-\class{steps-node}{\cssId{steps-node-14}{\dfrac{-1}{\sqrt{1-{\left(3x\right)}^{2}}}}}{\cdot}\class{steps-node}{\cssId{steps-node-15}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(3x\right)}}+3{\cdot}\class{steps-node}{\cssId{steps-node-12}{-\sin\left(4x\right)}}{\cdot}\class{steps-node}{\cssId{steps-node-13}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(4x\right)}}+2{\cdot}\class{steps-node}{\cssId{steps-node-10}{\cos\left(5x\right)}}{\cdot}\class{steps-node}{\cssId{steps-node-11}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(5x\right)}}+\class{steps-node}{\cssId{steps-node-8}{\dfrac{1}{1-{x}^{3}}}}{\cdot}\class{steps-node}{\cssId{steps-node-9}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(1-{x}^{3}\right)}}$

$=2{\cdot}\class{steps-node}{\cssId{steps-node-22}{5}}{\cdot}\cos\left(5x\right)-3{\cdot}\class{steps-node}{\cssId{steps-node-21}{4}}{\cdot}\sin\left(4x\right)+\class{steps-node}{\cssId{steps-node-18}{\left(\class{steps-node}{\cssId{steps-node-19}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{2}\right)}}-2\right)}}{\cdot}{\mathrm{e}}^{{x}^{2}-2x+1}+\dfrac{\class{steps-node}{\cssId{steps-node-23}{-\class{steps-node}{\cssId{steps-node-24}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({x}^{3}\right)}}}}}{1-{x}^{3}}+\dfrac{\class{steps-node}{\cssId{steps-node-20}{3}}}{\sqrt{1-9{x}^{2}}}$

$=10{\cdot}\cos\left(5x\right)-12{\cdot}\sin\left(4x\right)+\left(\class{steps-node}{\cssId{steps-node-25}{2}}\class{steps-node}{\cssId{steps-node-26}{x}}-2\right){\cdot}{\mathrm{e}}^{{x}^{2}-2x+1}-\dfrac{\class{steps-node}{\cssId{steps-node-27}{3}}\class{steps-node}{\cssId{steps-node-28}{{x}^{2}}}}{1-{x}^{3}}+\dfrac{3}{\sqrt{1-9{x}^{2}}}$