Question

1. the composite function ( z = x+sinleft(\frac{y}{x}

ight) ) with ( y = x^{2} ) has derivatives ( z_{x} ) and ( \frac{dz}{dx} ) which are respectively
a. ( z_{x} = 1+\frac{y}{x^{2}}cosleft(\frac{y}{x}
ight) ), ( \frac{dz}{dx}=1 - cos x ).
b. ( z_{x} = 1-\frac{y}{x^{2}}cosleft(\frac{y}{x}
ight) ), ( \frac{dz}{dx}=1 - cos x ).
c. ( z_{x} = 1+\frac{y}{x^{2}}cosleft(\frac{y}{x}
ight) ), ( \frac{dz}{dx}=1+cos x ).
d. ( z_{x} = 1-\frac{y}{x^{2}}cosleft(\frac{y}{x}
ight) ), ( \frac{dz}{dx}=1+cos x ).

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