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Pochodna funkcji (1+tg(2x))^(1/x)

$f\left(g, t, x\right) =$	${\left(2gtx+1\right)}^{\frac{1}{x}}$
$\dfrac{\mathrm{d}}{\mathrm{d}x}f\left(g, t, x\right) =$	$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({\left(2gtx+1\right)}^{\frac{1}{x}}\right)}}$ $=\class{steps-node}{\cssId{steps-node-2}{{\left(2gtx+1\right)}^{\frac{1}{x}}}}{\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\ln\left(2gtx+1\right){\cdot}\dfrac{1}{x}\right)}}$ $=\class{steps-node}{\cssId{steps-node-4}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{\ln\left(2gtx+1\right)}{x}\right)}}{\cdot}{\left(2gtx+1\right)}^{\frac{1}{x}}$ $=\dfrac{\class{steps-node}{\cssId{steps-node-7}{x{\cdot}\class{steps-node}{\cssId{steps-node-6}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\ln\left(2gtx+1\right)\right)}}}}-\class{steps-node}{\cssId{steps-node-9}{\class{steps-node}{\cssId{steps-node-8}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x\right)}}{\cdot}\ln\left(2gtx+1\right)}}}{\class{steps-node}{\cssId{steps-node-5}{{x}^{2}}}}{\cdot}{\left(2gtx+1\right)}^{\frac{1}{x}}$ $=\dfrac{{\left(2gtx+1\right)}^{\frac{1}{x}}{\cdot}\left(\class{steps-node}{\cssId{steps-node-10}{\dfrac{1}{2gtx+1}}}{\cdot}\class{steps-node}{\cssId{steps-node-11}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(2gtx+1\right)}}{\cdot}x-\class{steps-node}{\cssId{steps-node-12}{1}}{\cdot}\ln\left(2gtx+1\right)\right)}{{x}^{2}}$ $=\dfrac{{\left(2gtx+1\right)}^{\frac{1}{x}}{\cdot}\left(\dfrac{\class{steps-node}{\cssId{steps-node-13}{2gt}}x}{2gtx+1}-\ln\left(2gtx+1\right)\right)}{{x}^{2}}$ Uproszczony wynik: $={\left(2gtx+1\right)}^{\frac{1}{x}}{\cdot}\left(\dfrac{2gt}{x{\cdot}\left(2gtx+1\right)}-\dfrac{\ln\left(2gtx+1\right)}{{x}^{2}}\right)$

$f\left(g, t, x\right) =$

${\left(2gtx+1\right)}^{\frac{1}{x}}$

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

$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({\left(2gtx+1\right)}^{\frac{1}{x}}\right)}}$

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

$=\class{steps-node}{\cssId{steps-node-4}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\dfrac{\ln\left(2gtx+1\right)}{x}\right)}}{\cdot}{\left(2gtx+1\right)}^{\frac{1}{x}}$

$=\dfrac{\class{steps-node}{\cssId{steps-node-7}{x{\cdot}\class{steps-node}{\cssId{steps-node-6}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\ln\left(2gtx+1\right)\right)}}}}-\class{steps-node}{\cssId{steps-node-9}{\class{steps-node}{\cssId{steps-node-8}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x\right)}}{\cdot}\ln\left(2gtx+1\right)}}}{\class{steps-node}{\cssId{steps-node-5}{{x}^{2}}}}{\cdot}{\left(2gtx+1\right)}^{\frac{1}{x}}$

$=\dfrac{{\left(2gtx+1\right)}^{\frac{1}{x}}{\cdot}\left(\class{steps-node}{\cssId{steps-node-10}{\dfrac{1}{2gtx+1}}}{\cdot}\class{steps-node}{\cssId{steps-node-11}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(2gtx+1\right)}}{\cdot}x-\class{steps-node}{\cssId{steps-node-12}{1}}{\cdot}\ln\left(2gtx+1\right)\right)}{{x}^{2}}$

$=\dfrac{{\left(2gtx+1\right)}^{\frac{1}{x}}{\cdot}\left(\dfrac{\class{steps-node}{\cssId{steps-node-13}{2gt}}x}{2gtx+1}-\ln\left(2gtx+1\right)\right)}{{x}^{2}}$

Uproszczony wynik:

$={\left(2gtx+1\right)}^{\frac{1}{x}}{\cdot}\left(\dfrac{2gt}{x{\cdot}\left(2gtx+1\right)}-\dfrac{\ln\left(2gtx+1\right)}{{x}^{2}}\right)$