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Pochodna funkcji (sinx)^tgx

$f\left(g, t, x\right) =$	$gx{\cdot}{\left(\sin\left(x\right)\right)}^{t}$
$\dfrac{\mathrm{d}\left(f\left(g, t, x\right)\right)}{\mathrm{d}x} =$	$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(gx{\cdot}{\left(\sin\left(x\right)\right)}^{t}\right)}}$ $=\class{steps-node}{\cssId{steps-node-2}{g{\cdot}\class{steps-node}{\cssId{steps-node-3}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x{\cdot}{\left(\sin\left(x\right)\right)}^{t}\right)}}}}$ $=g{\cdot}\left(\class{steps-node}{\cssId{steps-node-5}{\class{steps-node}{\cssId{steps-node-4}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(x\right)}}{\cdot}{\left(\sin\left(x\right)\right)}^{t}}}+\class{steps-node}{\cssId{steps-node-7}{x{\cdot}\class{steps-node}{\cssId{steps-node-6}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left({\left(\sin\left(x\right)\right)}^{t}\right)}}}}\right)$ $=g{\cdot}\left(\class{steps-node}{\cssId{steps-node-8}{1}}{\cdot}{\left(\sin\left(x\right)\right)}^{t}+\class{steps-node}{\cssId{steps-node-9}{t}}{\cdot}\class{steps-node}{\cssId{steps-node-10}{{\left(\sin\left(x\right)\right)}^{t-1}}}{\cdot}\class{steps-node}{\cssId{steps-node-11}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sin\left(x\right)\right)}}{\cdot}x\right)$ $=g{\cdot}\left({\left(\sin\left(x\right)\right)}^{t}+t{\cdot}\class{steps-node}{\cssId{steps-node-12}{\cos\left(x\right)}}{\cdot}x{\cdot}{\left(\sin\left(x\right)\right)}^{t-1}\right)$ $=g{\cdot}\left({\left(\sin\left(x\right)\right)}^{t}+tx{\cdot}\cos\left(x\right){\cdot}{\left(\sin\left(x\right)\right)}^{t-1}\right)$ Wynik alternatywny: $=g{\cdot}{\left(\sin\left(x\right)\right)}^{t}+gtx{\cdot}\cos\left(x\right){\cdot}{\left(\sin\left(x\right)\right)}^{t-1}$

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

$gx{\cdot}{\left(\sin\left(x\right)\right)}^{t}$

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

$\class{steps-node}{\cssId{steps-node-1}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(gx{\cdot}{\left(\sin\left(x\right)\right)}^{t}\right)}}$

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

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

$=g{\cdot}\left(\class{steps-node}{\cssId{steps-node-8}{1}}{\cdot}{\left(\sin\left(x\right)\right)}^{t}+\class{steps-node}{\cssId{steps-node-9}{t}}{\cdot}\class{steps-node}{\cssId{steps-node-10}{{\left(\sin\left(x\right)\right)}^{t-1}}}{\cdot}\class{steps-node}{\cssId{steps-node-11}{\tfrac{\mathrm{d}}{\mathrm{d}x}\kern-.25em\left(\sin\left(x\right)\right)}}{\cdot}x\right)$

$=g{\cdot}\left({\left(\sin\left(x\right)\right)}^{t}+t{\cdot}\class{steps-node}{\cssId{steps-node-12}{\cos\left(x\right)}}{\cdot}x{\cdot}{\left(\sin\left(x\right)\right)}^{t-1}\right)$

$=g{\cdot}\left({\left(\sin\left(x\right)\right)}^{t}+tx{\cdot}\cos\left(x\right){\cdot}{\left(\sin\left(x\right)\right)}^{t-1}\right)$

Wynik alternatywny:

$=g{\cdot}{\left(\sin\left(x\right)\right)}^{t}+gtx{\cdot}\cos\left(x\right){\cdot}{\left(\sin\left(x\right)\right)}^{t-1}$