1. let ( z = ue^{v} ) where ( u = u(x,y) ), ( v = v(x,y) ). the partial derivative ( z_{x} ) is defined…

1. let ( z = ue^{v} ) where ( u = u(x,y) ), ( v = v(x,y) ). the partial derivative ( z_{x} ) is defined by\na. ( z_{x} = e^{v}u_{x}+ue^{v}v_{x} ).\nb. ( z_{x} = ue^{v}u_{x}+e^{v}v_{x} ).\nc. ( z_{x} = v_{x}+e^{v}u_{x} ).\nd. ( z_{x} = v_{x}e^{v}u_{x} ).

1. let ( z = ue^{v} ) where ( u = u(x,y) ), ( v = v(x,y) ). the partial derivative ( z_{x} ) is defined by\na. ( z_{x} = e^{v}u_{x}+ue^{v}v_{x} ).\nb. ( z_{x} = ue^{v}u_{x}+e^{v}v_{x} ).\nc. ( z_{x} = v_{x}+e^{v}u_{x} ).\nd. ( z_{x} = v_{x}e^{v}u_{x} ).

Answer

Explanation:

Step1: Apply the product rule for partial derivatives

The product rule for partial derivatives states that if (z = f(x,y)g(x,y)), then (\frac{\partial z}{\partial x}=\frac{\partial f}{\partial x}g(x,y)+f(x,y)\frac{\partial g}{\partial x}). Here, (f = u) and (g = e^{v}). So, (z_{x}'=\frac{\partial(u)}{\partial x}e^{v}+u\frac{\partial(e^{v})}{\partial x}).

Step2: Apply the chain rule for (\frac{\partial(e^{v})}{\partial x})

The chain rule for partial derivatives: if (y = h(k(x,y))), then (\frac{\partial y}{\partial x}=h^{\prime}(k(x,y))\frac{\partial k}{\partial x}). For (y = e^{v}), (h(t)=e^{t}), (k = v(x,y)), so (\frac{\partial(e^{v})}{\partial x}=e^{v}v_{x}'). Substituting back into the expression from Step1: (z_{x}'=e^{v}u_{x}'+ue^{v}v_{x}')

Answer:

A. (z_{x}' = e^{v}u_{x}'+ue^{v}v_{x}')