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

2. the composite function ( z = x+sinleft(\frac{y}{x}\right) ) with ( y = x^{2} ) has derivatives ( z_{x}^{prime} ) and ( \frac{dz}{dx} ) which are respectively\na. ( z_{x}^{prime}=1+\frac{y}{x^{2}}cosleft(\frac{y}{x}\right) ), ( \frac{dz}{dx}=1 - cos x ).\nb. ( z_{x}^{prime}=1-\frac{y}{x^{2}}cosleft(\frac{y}{x}\right) ), ( \frac{dz}{dx}=1 - cos x ).\nc. ( z_{x}^{prime}=1+\frac{y}{x^{2}}cosleft(\frac{y}{x}\right) ), ( \frac{dz}{dx}=1+cos x ).\nd. ( z_{x}^{prime}=1-\frac{y}{x^{2}}cosleft(\frac{y}{x}\right) ), ( \frac{dz}{dx}=1+cos x ).

2. the composite function ( z = x+sinleft(\frac{y}{x}\right) ) with ( y = x^{2} ) has derivatives ( z_{x}^{prime} ) and ( \frac{dz}{dx} ) which are respectively\na. ( z_{x}^{prime}=1+\frac{y}{x^{2}}cosleft(\frac{y}{x}\right) ), ( \frac{dz}{dx}=1 - cos x ).\nb. ( z_{x}^{prime}=1-\frac{y}{x^{2}}cosleft(\frac{y}{x}\right) ), ( \frac{dz}{dx}=1 - cos x ).\nc. ( z_{x}^{prime}=1+\frac{y}{x^{2}}cosleft(\frac{y}{x}\right) ), ( \frac{dz}{dx}=1+cos x ).\nd. ( z_{x}^{prime}=1-\frac{y}{x^{2}}cosleft(\frac{y}{x}\right) ), ( \frac{dz}{dx}=1+cos x ).

Answer

Explanation:

Step1: Find (z'_x)

Use 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, use the chain rule. Let (u=\frac{y}{x}), then (\frac{\partial\sin(u)}{\partial x}=\cos(u)\cdot\frac{\partial u}{\partial x}). Since (u = \frac{y}{x}), (\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})

Use 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(\frac{x^{2}}{x})+\frac{1}{x}\cos(\frac{x^{2}}{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)