Question

Calculate dy/dx if Ln (x + y) = ex/y

A: e^x/y xy + e^x/y y^2 - y^2
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e^x/y xy + e^x/y x^2 + y2


B: e^x/y xy + e^x/y y^2 - y^2
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e^x/y xy - e^x/y x^2 + y^2


C: e^x/y xy + e^x/y y^2 + y^2
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e^x/y xy - e^x/y x^2 + y^2

D: e^x/y xy + e^x/y y^2 + y^2
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e^x/y xy + e^x/y x^2 + y^2

Answer 1

we are given

`ln(x+y)=e^{(x)/(y) }`

Since, we have to solve for dy/dx

so, we will take derivative with respect to x on both sides

`(d)/(dx)\left(\ln \left(x+y\right)\right)=(d)/(dx)\left(e^{(x)/(y)}\right)`

Left side:

`(d)/(dx)\left(\ln \left(x+y\right)\right)`

we can use chain rule

u=x+y

`=(d)/(du)\left(\ln \left(u\right)\right)(d)/(dx)\left(x+y\right)`

`=(1)/(u)\left(1+(d)/(dx)\left(y\right)\right)`

now, we can plug back

u=x+y

`=(1)/(x+y)\left(1+(d)/(dx)\left(y\right)\right)`

`=(1+(d)/(dx)\left(y\right))/(x+y)`

Right side:

`(d)/(dx)\left(e^{(x)/(y)}\right)`

`=e^{(x)/(y)}(y-x(d)/(dx)\left(y\right))/(y^2)`

`=\frac{e^{(x)/(y)}\left(y-x(d)/(dx)\left(y\right)\right)}{y^2}`

now, we can set them equal

`(1+(d)/(dx)\left(y\right))/(x+y)=\frac{e^{(x)/(y)}\left(y-x(d)/(dx)\left(y\right)\right)}{y^2}`

now, we can solve for dy/dx

we get

`(dy)/(dx)=\frac{yxe^{(x)/(y)}+y^2e^{(x)/(y)}-y^2}{y^2+yxe^{(x)/(y)}+x^2e^{(x)/(y)}}`..............Answer

Answer 2

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]