$L e t r = x^{2} + y - z a n d z^{3} - x y + y z + y^{3} = 1$ Assume that x and y are independent variables, At (x, y, z) = (2,-1, 1), the value (correct to two decimal places) of $\frac{\partial r}{\partial x}$ is
4.5

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Solution

The correct option is A 4.5
Given that $z^{3} - x y + y z + y^{3} = 1$

and x, y are independent variables so we can take z = f(x, y)

i.e, Differentiating (1) partially w.r.t 'x'

$3 z^{2} \frac{\partial z}{\partial x} - y + y \frac{\partial z}{\partial x} + 0 = 0$

$\Rightarrow \frac{\partial z}{\partial x} = \frac{y}{3 z^{2} + y}$

Now r = $x^{2} + y - z$

$\frac{\partial r}{\partial x} = 2 x + 0 - \frac{\partial z}{\partial x}$

= 2x - $\frac{y}{3 z^{2} + y}$

${(\frac{\partial r}{\partial x})}_{(2, - 1, 1)} = 2 \times 2 - \frac{- 1}{3 (1)^{2} - 1}$

= 4 + $\frac{1}{2} = 4.5$

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