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 \(z_{x}'\)

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

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Answer:

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