# Differentiate implicitly to find the first partial derivatives of Z = e^x sin(y + z) given that z = f(x,y) without using any theorem.

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### 1 Answer

You need to differentiate the function with respect to x, keeping y constant, using chain rule, such that:

`(del z)/(del x) = (e^x)'*sin(y + z) + (e^x)*(sin(y + z))'`

`(del z)/(del x) = (e^x)*sin(y + z) + (e^x)*(cos(y + z))*(0 + (del z)/(del x))`

`(del z)/(del x) = (e^x)*sin(y + z) + (e^x)*(cos(y + z))*((del z)/(del x))`

You need to move the terms that contain `(del z)/(del x)` to the left side such that:

`(del z)/(del x)-(e^x)*(cos(y + z))*((del z)/(del x)) = (e^x)*sin(y + z)`

Factoring out `(del z)/(del x)` yields:

`(del z)/(del x)(1 - (e^x)*(cos(y + z))) = (e^x)*sin(y + z)`

`(del z)/(del x) = ((e^x)*sin(y + z))/(1 - (e^x)*(cos(y + z))) `

You need to differentiate the function with respect to y, keeping x constant, using chain rule, such that:

`(del z)/(del y) = (e^x)'*sin(y + z) + (e^x)*(sin(y + z))'`

`(del z)/(del y) = 0*sin(y + z) + (e^x)*(cos(y + z))*(1 + (del z)/(del y))`

`(del z)/(del y) = (e^x)*(cos(y + z)) + (e^x)*(cos(y + z))*((del z)/(del y))`

You need to move the terms that contain `(del z)/(del y)` to the left side such that:

`(del z)/(del y) - (e^x)*(cos(y + z))*((del z)/(del y)) = (e^x)*(cos(y + z))`

Factoring out `(del z)/(del y)` yields:

`(del z)/(del y)(1 -(e^x)*(cos(y + z))) = (e^x)*(cos(y + z))`

`(del z)/(del y) = ((e^x)*(cos(y + z)))/(1 - (e^x)*(cos(y + z))) `

Hence, evaluating partial derivatives yields `(del z)/(del x) = ((e^x)*sin(y + z))/(1 - (e^x)*(cos(y + z)))` and `(del z)/(del y) = ((e^x)*(cos(y + z)))/(1 - (e^x)*(cos(y + z))).`