What is (\partial)/(\partial x)(x^2ye^{x-z}) ?

The answer to (\partial)/(\partial x)(x^2ye^{x-z}) is y(2xe^{x-z}+e^{x-z}x^2)

Popular >

(\partial)/(\partial x)(x^2ye^{x-z})

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

y (2 x e^{x - z} + e^{x - z} x^{2})

Solution steps

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

Treat

y, z

as constants

Take the constant out:

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

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

Apply the Product Rule:

(f \cdot g)^{'} = f^{'} \cdot g + f \cdot g^{'}

f = x^{2}, g = e^{x - z}

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

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

\frac{\partial}{\partial x} (e^{x - z}) = e^{x - z}

= y (2 x e^{x - z} + e^{x - z} x^{2})

\frac{1}{1 + s} \cdot \frac{8}{s}

\int \frac{e ^{u}}{( 1 - e ^{u} ) ^{2}} d u

\frac{d y}{d x} = y^{2} (4 - e^{x})

3 y^{^{''}} + 5 y^{^{'}} - 12 y = 0

\frac{d}{d y} (- 2 e^{x} + 4 x cos (y))

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