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Question Number 93477 by Ar Brandon last updated on 13/May/20

Differentiate completely;  1\ f(x,y)=x^2 +xy^2 +siny  2\ f(x,y)=e^(x^2 +y^2 )   3\ f(x,y,z)=tan(3x−y)+6^(y+2)

$$\mathrm{Differentiate}\:\mathrm{completely}; \\ $$$$\mathrm{1}\backslash\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{x}^{\mathrm{2}} +\mathrm{xy}^{\mathrm{2}} +\mathrm{siny} \\ $$$$\mathrm{2}\backslash\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)=\mathrm{e}^{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} } \\ $$$$\mathrm{3}\backslash\:\mathrm{f}\left(\mathrm{x},\mathrm{y},\mathrm{z}\right)=\mathrm{tan}\left(\mathrm{3x}−\mathrm{y}\right)+\mathrm{6}^{\mathrm{y}+\mathrm{2}} \\ $$

Answered by Rio Michael last updated on 13/May/20

partial derivative     (∂f/∂x) = 2x + y^2   , (∂f/∂y) = 2xy + cos y

$$\mathrm{partial}\:\mathrm{derivative}\: \\ $$$$\:\:\frac{\partial{f}}{\partial{x}}\:=\:\mathrm{2}{x}\:+\:{y}^{\mathrm{2}} \:\:,\:\frac{\partial{f}}{\partial{y}}\:=\:\mathrm{2}{xy}\:+\:\mathrm{cos}\:{y} \\ $$

Commented by Ar Brandon last updated on 13/May/20

Hi thanks for your reply. But actually  I′m looking for the total derivative.  Do you have an idea on how to go about?

$$\mathrm{Hi}\:\mathrm{thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{reply}.\:\mathrm{But}\:\mathrm{actually} \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{the}\:\mathrm{total}\:\mathrm{derivative}. \\ $$$$\mathrm{Do}\:\mathrm{you}\:\mathrm{have}\:\mathrm{an}\:\mathrm{idea}\:\mathrm{on}\:\mathrm{how}\:\mathrm{to}\:\mathrm{go}\:\mathrm{about}? \\ $$

Answered by john santu last updated on 13/May/20

(1) let z = f(x,y) = x^2 +xy^2 +sin y  dz = (∂z/∂x).dx + (∂z/∂y).dy   dz = (2x+y^2 )dx +(2xy+cos y)dy

$$\left(\mathrm{1}\right)\:\mathrm{let}\:\mathrm{z}\:=\:\mathrm{f}\left(\mathrm{x},\mathrm{y}\right)\:=\:\mathrm{x}^{\mathrm{2}} +\mathrm{xy}^{\mathrm{2}} +\mathrm{sin}\:\mathrm{y} \\ $$$$\mathrm{dz}\:=\:\frac{\partial\mathrm{z}}{\partial\mathrm{x}}.\mathrm{dx}\:+\:\frac{\partial\mathrm{z}}{\partial\mathrm{y}}.\mathrm{dy}\: \\ $$$$\mathrm{dz}\:=\:\left(\mathrm{2x}+\mathrm{y}^{\mathrm{2}} \right)\mathrm{dx}\:+\left(\mathrm{2xy}+\mathrm{cos}\:\mathrm{y}\right)\mathrm{dy} \\ $$$$ \\ $$

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