Respuesta :
Given that ln(x+y)=e^(x/y)
In this case we use the quotient rule on the right hand side where we have an exponent.
1/(x+y)[1+dy/dx]=e^(x/y)[[(y)-x(dy/dx)]/[y^2]]
opening the parenthesis on the RHS
1/(x+y)[1+dy/dx]=e^(x/y)[1/y-(x/y^2)(dy/dx)]
Expanding the left hand side and right hand side
1/(x+y)+(dy/dx)[1/(x+y)]=[e^(x/y)]/y-[e^(x/y)](x/y^2)(dy/dx)
next we put like terms together by moving all terms with dy/dx to the LHS and everything else to the RHS.
(dy/dx)[1/(x+y)]+[e^(x/y)](x/y^2)(dy/dx)=[e^(x/y)]/y-1/(x+y)
dy/dx[1/(x+y)+[e^(x/y)](x/y^2)]=[e^(x/y)]/y-1/(x+y)
we can replace e^(x/y) with ln(x+y)
dy/dx[1/(x+y)+[ln(x+y)](x/y^2)]=[ln(x+y)]/y-1/(x+y)
Multiplying both sides by (x+y)(y^2)
dy/dx[y^2+(x+y)ln(x+y)]=y(x+y)ln(x+y)-y^2
dy/dx=[y(x+y)ln(x+y)-y^2]/[y^2+(x+y)ln(x+y)]
This can be written as:
dy/dx=[y(x+y)e^(x/y)-y^2]/[y^2+(x+y)e^(x/y)]
the answer is A]
In this case we use the quotient rule on the right hand side where we have an exponent.
1/(x+y)[1+dy/dx]=e^(x/y)[[(y)-x(dy/dx)]/[y^2]]
opening the parenthesis on the RHS
1/(x+y)[1+dy/dx]=e^(x/y)[1/y-(x/y^2)(dy/dx)]
Expanding the left hand side and right hand side
1/(x+y)+(dy/dx)[1/(x+y)]=[e^(x/y)]/y-[e^(x/y)](x/y^2)(dy/dx)
next we put like terms together by moving all terms with dy/dx to the LHS and everything else to the RHS.
(dy/dx)[1/(x+y)]+[e^(x/y)](x/y^2)(dy/dx)=[e^(x/y)]/y-1/(x+y)
dy/dx[1/(x+y)+[e^(x/y)](x/y^2)]=[e^(x/y)]/y-1/(x+y)
we can replace e^(x/y) with ln(x+y)
dy/dx[1/(x+y)+[ln(x+y)](x/y^2)]=[ln(x+y)]/y-1/(x+y)
Multiplying both sides by (x+y)(y^2)
dy/dx[y^2+(x+y)ln(x+y)]=y(x+y)ln(x+y)-y^2
dy/dx=[y(x+y)ln(x+y)-y^2]/[y^2+(x+y)ln(x+y)]
This can be written as:
dy/dx=[y(x+y)e^(x/y)-y^2]/[y^2+(x+y)e^(x/y)]
the answer is A]
we are given
[tex]ln(x+y)=e^{\frac{x}{y} }[/tex]
Since, we have to solve for dy/dx
so, we will take derivative with respect to x on both sides
[tex]\frac{d}{dx}\left(\ln \left(x+y\right)\right)=\frac{d}{dx}\left(e^{\frac{x}{y}}\right)[/tex]
Left side:
[tex]\frac{d}{dx}\left(\ln \left(x+y\right)\right)[/tex]
we can use chain rule
u=x+y
[tex]=\frac{d}{du}\left(\ln \left(u\right)\right)\frac{d}{dx}\left(x+y\right)[/tex]
[tex]=\frac{1}{u}\left(1+\frac{d}{dx}\left(y\right)\right)[/tex]
now, we can plug back
u=x+y
[tex]=\frac{1}{x+y}\left(1+\frac{d}{dx}\left(y\right)\right)[/tex]
[tex]=\frac{1+\frac{d}{dx}\left(y\right)}{x+y}[/tex]
Right side:
[tex]\frac{d}{dx}\left(e^{\frac{x}{y}}\right)[/tex]
[tex]=e^{\frac{x}{y}}\frac{y-x\frac{d}{dx}\left(y\right)}{y^2}[/tex]
[tex]=\frac{e^{\frac{x}{y}}\left(y-x\frac{d}{dx}\left(y\right)\right)}{y^2}[/tex]
now, we can set them equal
[tex]\frac{1+\frac{d}{dx}\left(y\right)}{x+y}=\frac{e^{\frac{x}{y}}\left(y-x\frac{d}{dx}\left(y\right)\right)}{y^2}[/tex]
now, we can solve for dy/dx
we get
[tex]\frac{dy}{dx}=\frac{yxe^{\frac{x}{y}}+y^2e^{\frac{x}{y}}-y^2}{y^2+yxe^{\frac{x}{y}}+x^2e^{\frac{x}{y}}}[/tex]..............Answer
