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

The answer to (\partial)/(\partial x)(y^2e^{xy^2+z^2}) is e^{y^2x+z^2}y^4

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(\partial)/(\partial x)(y^2e^{xy^2+z^2})

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

e^{y^{2} x + z^{2}} y^{4}

Solution steps

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

Treat

y, z

as constants

Take the constant out:

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

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

Apply the chain rule:

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

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

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

= y^{2} e^{x y^{2} + z^{2}} y^{2}

Simplify

y^{2} e^{x y^{2} + z^{2}} y^{2} : e^{y^{2} x + z^{2}} y^{4}

= e^{y^{2} x + z^{2}} y^{4}

y^{^{'}} = x^{2} e^{2 y}, y (- 1) = - 1

\frac{d}{d t} (3 cos (t) - 2 e^{2 t})

(- 1, 0), (0, - 2)

(x^{2} + 3 x y + y^{2}) d x - x^{2} d y = 0

cos (x), - \frac{π}{2}

What is (\partial)/(\partial x)(y^2e^{xy^2+z^2}) ?
The answer to (\partial)/(\partial x)(y^2e^{xy^2+z^2}) is e^{y^2x+z^2}y^4