gdzie jesteś: Start / Pochodne / Pochodne funkcji trygonometrycznych / Pochodna funkcji ln(1+sinx)

Pochodna funkcji ln(1+sinx)

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

$f\left(x\right) =$

$\ln\left(\sin\left(x\right)+1\right)$

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

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

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

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

$=\dfrac{\class{steps-node}{\cssId{steps-node-5}{\cos\left(x\right)}}}{\sin\left(x\right)+1}$