Pochodna funkcji (sinx/cosx)+x

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

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

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

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

$=\dfrac{\class{steps-node}{\cssId{steps-node-9}{\cos\left(x\right)}}{\cdot}\cos\left(x\right)-\class{steps-node}{\cssId{steps-node-10}{-\sin\left(x\right)}}{\cdot}\sin\left(x\right)}{{\left(\cos\left(x\right)\right)}^{2}}+1$

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

Uproszczony wynik:

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

Podziel się rozwiązaniem:

Wybrane przykłady