What is (\partial)/(\partial x)(we^{x^2+y^2}ln(z)) ?

The answer to (\partial)/(\partial x)(we^{x^2+y^2}ln(z)) is wln(z)e^{x^2+y^2}*2x

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(\partial)/(\partial x)(we^{x^2+y^2}ln(z))

\frac{\partial}{\partial x} (w e^{x^{2} + y^{2}} ln (z))

w ln (z) e^{x^{2} + y^{2}} \cdot 2 x

Solution steps

\frac{\partial}{\partial x} (w e^{x^{2} + y^{2}} ln (z))

Treat

w, y, z

as constants

Take the constant out:

(a \cdot f)^{'} = a \cdot f^{'}

= w ln (z) \frac{\partial}{\partial x} (e^{x^{2} + y^{2}})

Apply the chain rule:

e^{x^{2} + y^{2}} \frac{\partial}{\partial x} (x^{2} + y^{2})

= e^{x^{2} + y^{2}} \frac{\partial}{\partial x} (x^{2} + y^{2})

\frac{\partial}{\partial x} (x^{2} + y^{2}) = 2 x

= w ln (z) e^{x^{2} + y^{2}} \cdot 2 x

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What is (\partial)/(\partial x)(we^{x^2+y^2}ln(z)) ?
The answer to (\partial)/(\partial x)(we^{x^2+y^2}ln(z)) is wln(z)e^{x^2+y^2}*2x