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Question Number 203448 by hardmath last updated on 19/Jan/24

y = sin3x−ln(3x+1)+3^(2x)   find:   y^′  = ?

$$\mathrm{y}\:=\:\mathrm{sin3x}−\mathrm{ln}\left(\mathrm{3x}+\mathrm{1}\right)+\mathrm{3}^{\mathrm{2}\boldsymbol{\mathrm{x}}} \\ $$$$\mathrm{find}:\:\:\:\mathrm{y}^{'} \:=\:? \\ $$

Commented by hardmath last updated on 19/Jan/24

  find the derivative of the function

$$ \\ $$find the derivative of the function

Answered by mr W last updated on 19/Jan/24

y′=3 cos 3x−(3/(3x+1))+3^(2x) ×2 ln 3

$${y}'=\mathrm{3}\:\mathrm{cos}\:\mathrm{3}{x}−\frac{\mathrm{3}}{\mathrm{3}{x}+\mathrm{1}}+\mathrm{3}^{\mathrm{2}{x}} ×\mathrm{2}\:\mathrm{ln}\:\mathrm{3} \\ $$

Commented by hardmath last updated on 19/Jan/24

thank you dear professor

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{dear}\:\mathrm{professor} \\ $$

Answered by MrGHK last updated on 19/Jan/24

y^′ =(d/dx)(3x).(d/du)(sin(u))−(d/dx)(3x+1).(d/du)(ln(u))+(d/dx)(2xln(3)).(d/du)e^u   y^′ =3cos(3x)−(3/(3x+1))+2ln(3)3^(2x)

$$\boldsymbol{\mathrm{y}}^{'} =\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{dx}}}\left(\mathrm{3}\boldsymbol{\mathrm{x}}\right).\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{du}}}\left(\boldsymbol{\mathrm{sin}}\left(\boldsymbol{\mathrm{u}}\right)\right)−\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{dx}}}\left(\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{1}\right).\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{du}}}\left(\boldsymbol{\mathrm{ln}}\left(\boldsymbol{\mathrm{u}}\right)\right)+\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{dx}}}\left(\mathrm{2}\boldsymbol{\mathrm{xln}}\left(\mathrm{3}\right)\right).\frac{\boldsymbol{\mathrm{d}}}{\boldsymbol{\mathrm{du}}}\boldsymbol{\mathrm{e}}^{\boldsymbol{\mathrm{u}}} \\ $$$$\boldsymbol{\mathrm{y}}^{'} =\mathrm{3}\boldsymbol{\mathrm{cos}}\left(\mathrm{3}\boldsymbol{\mathrm{x}}\right)−\frac{\mathrm{3}}{\mathrm{3}\boldsymbol{\mathrm{x}}+\mathrm{1}}+\mathrm{2}\boldsymbol{\mathrm{ln}}\left(\mathrm{3}\right)\mathrm{3}^{\mathrm{2}\boldsymbol{\mathrm{x}}} \\ $$

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