Question

dy/dx[e^(x/y)=4x-5y]

Answer 1

i am stuck with the e^(x/y) part

Answer 2

e^(x/y) ((y-xy')/(y^2)) = 4 - 5y'
I think you can solve for y' at this point. Make sure you bring all y' to one side, and eberything else to the other side; then factor out y'..and you'll be set.

Answer 3

The derivative of e^u is e^u du

so the derivative of e^(x/y) is e^(x/y) times the nderivative of x/y (use quotient rule).

Answer 4

The derivative of x/y will be (y-xy')/(y^2)

Answer 5

The derivative of y is just y'

Answer 6

i am lost

Answer 7

\[e ^{x/y}=4x-5y\]

Answer 8

You want the derivative yeah?

Answer 9

yup

Answer 10

\[\frac{ d }{ dx }(e^{x/y}) = \frac{ d }{ dx }(4x-5y)\]

\[e ^{\frac{ x }{ y }}\frac{ d }{ dx }(\frac{ x }{ y }) = \frac{ d }{ dx }(4x-5y) \] Chain rule


\[\frac{ \frac{ d }{ dx }(x)y-x \frac{ d }{ dx}(y) }{ y^2 }e ^{x/y}= \frac{ d }{ dx }(4x-5y)\] quotient rule


\[\frac{ e ^{x/y}(1y-xy'(x)) }{ y^2 } = \frac{ d }{ dx }(4x-5y) \]


\[\frac{ e ^{x/y}(y-xy'(x)) }{ y^2 }= 4\frac{ d }{ dx }(x)-5\frac{ d }{ dx }(y)\] Differentiate the sum term by term and factor out constants.


\[\frac{ e ^{x/y}(y-xy'(x)) }{ y^2 } = 4-5 y'(x)\]


\[\frac{ e^{x/y} }{ y }-\frac{ e ^{x/y}xy'(x) }{ y^2 } = 4-5y'(x)\] Expand


\[\frac{ e ^{x/y} }{ y }+5y'(x) - \frac{ e ^{x/y}xy'(x) }{ y^2 } = 4\]
Add 5y'(x) to both sides


\[(5-\frac{ e ^{x/y}x }{ y^2 })y'(x) = 4 - \frac{ e ^{x/y} }{ y }\]
Collect left hand side for y'(x).


\[y'(x) = \frac{ 4-\frac{ e ^{x/y} }{ y } }{ 5-\frac{ e ^{x/y}x }{ y^2 } }\]