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

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

Answer

Explanation:

Step1: Find (z_{x}^{\prime})

Use the sum rule and the chain rule. The derivative of (x) with respect to (x) is (1). For (\sin(\frac{y}{x})), let (u = \frac{y}{x}), then (\frac{\partial}{\partial x}\sin(u)=\cos(u)\cdot\frac{\partial u}{\partial x}). (\frac{\partial u}{\partial x}=\frac{-y}{x^{2}}) (using the quotient rule (\frac{\partial}{\partial x}(\frac{a}{b})=\frac{-a}{b^{2}}) where (a = y) and (b=x)). So (z_{x}^{\prime}=1+\cos(\frac{y}{x})\cdot\frac{-y}{x^{2}}=1-\frac{y}{x^{2}}\cos(\frac{y}{x})).

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

Substitute (y = x^{2}) into (z=x+\sin(\frac{y}{x})), we get (z=x+\sin(x)). Using the sum rule (\frac{d}{dx}(u + v)=\frac{du}{dx}+\frac{dv}{dx}), where (u=x) and (v = \sin(x)). The derivative of (x) with respect to (x) is (1), and the derivative of (\sin(x)) with respect to (x) is (\cos(x)). So (\frac{dz}{dx}=1+\cos(x)).

Answer:

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