Question

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

a. ( z_{x} = e^{v}u_{x}+ue^{v}v_{x} ).
b. ( z_{x} = ue^{v}u_{x}+e^{v}v_{x} ).
c. ( z_{x} = v_{x}+e^{v}u_{x} ).
d. ( z_{x} = v_{x}e^{v}u_{x} ).

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