2. the composite function ( z = x+sinleft(\frac{y}{x}\right) ) with ( y = x^{2} ) has derivatives ( z_{x} )…

Explanation:

Step1: Find (z_{x}')

Using the partial - derivative formula. For (z = x+\sin(\frac{y}{x})), the partial derivative of (x) with respect to (x) is (1). For the second term, using the chain rule (\frac{\partial}{\partial x}\sin(u)=\cos(u)\cdot\frac{\partial u}{\partial x}), where (u = \frac{y}{x}). Then (\frac{\partial u}{\partial x}=-\frac{y}{x^{2}}). So (z_{x}'=1+\cos(\frac{y}{x})\cdot(-\frac{y}{x^{2}})=1-\frac{y}{x^{2}}\cos(\frac{y}{x})).

Step2: Find (\frac{dz}{dx})

Using the chain rule (\frac{dz}{dx}=\frac{\partial z}{\partial x}+\frac{\partial z}{\partial y}\cdot\frac{dy}{dx}). We know (\frac{\partial z}{\partial x}=1-\frac{y}{x^{2}}\cos(\frac{y}{x})), (\frac{\partial z}{\partial y}=\frac{1}{x}\cos(\frac{y}{x})), and (\frac{dy}{dx} = 2x). Substitute (y = x^{2}) into (\frac{dz}{dx}): [ \begin{align*} \frac{dz}{dx}&=1-\frac{x^{2}}{x^{2}}\cos(x)+\frac{1}{x}\cos(x)\cdot2x\ &=1 - \cos(x)+2\cos(x)\ &=1+\cos(x) \end{align*} ]

Answer:

D. (z_{x}' = 1-\frac{y}{x^{2}}\cos(\frac{y}{x})), (\frac{dz}{dx}=1+\cos x)