Find dz/dy at (1, ln 2, ln 3) if z(x, y) is defined by the equation xe^y+ye^z+2lnx-2-3ln2=0.

## Find dz/dy at (1, ln 2, ln 3) if z(x, y)?

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# 2 Answers

Differentiate both sides of xe^y+ye^z+2lnx-2-3ln2 = 0 with respect to y,

xe^y + e^z + ye^z ∂z/∂y = 0

Plug in (1, ln 2, ln 3),

2 + 3 + 3ln2 ∂z/∂y = 0

∂z/∂y = -5/(3ln2) <==Answer

You still can find the partial derivative ∂z/∂y even if z = ln 3 is not given to you. You can notice, if you substitute x=1 and y = ln2 into the equation, you will get z = ln 3 (you can express exp(z) via x and y and take natural logarithm of the function). You will come up with the function

z = -lny + ln[(ln(8e^{2}/x ^{
2}) - x e ^{y}]

Take the partial derivative

∂z/∂y = -1/y - xe^{ y}/[(ln(8 e^{2}/x
^{2}) -x e^{ y}]

If you substitute x =1 and y = ln2 you will get ∂z/∂y = - 5/(3ln2)