Pochodna funkcji xln(sqr(x))

$f\left(q, r, s, x\right) =$ $x{\cdot}\left(\ln\left(x\right)+\ln\left(s\right)+\ln\left(r\right)+\ln\left(q\right)\right)$

Note: Your input has been rewritten/simplified.

$\dfrac{\mathrm{d}\left(f\left(q, r, s, x\right)\right)}{\mathrm{d}x} =$

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

$=\class{steps-node}{\cssId{steps-node-3}{\class{steps-node}{\cssId{steps-node-2}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x\right)}}{\cdot}\left(\ln\left(x\right)+\ln\left(s\right)+\ln\left(r\right)+\ln\left(q\right)\right)}}+\class{steps-node}{\cssId{steps-node-5}{x{\cdot}\class{steps-node}{\cssId{steps-node-4}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\ln\left(x\right)+\ln\left(s\right)+\ln\left(r\right)+\ln\left(q\right)\right)}}}}$

$=\class{steps-node}{\cssId{steps-node-6}{1}}{\cdot}\left(\ln\left(x\right)+\ln\left(s\right)+\ln\left(r\right)+\ln\left(q\right)\right)+\class{steps-node}{\cssId{steps-node-7}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\ln\left(x\right)\right)}}{\cdot}x$

$=\ln\left(x\right)+\class{steps-node}{\cssId{steps-node-8}{\dfrac{1}{x}}}{\cdot}x+\ln\left(s\right)+\ln\left(r\right)+\ln\left(q\right)$

$=\ln\left(x\right)+\ln\left(s\right)+\ln\left(r\right)+\ln\left(q\right)+1$

Podziel się rozwiązaniem:

Wybrane przykłady