First slide

Derivatives

Question

$If e^{y} + x y = e, the ordered pair (\frac{d y}{d x}, \frac{d^{2} y}{d x^{2}}) at x = 0 is equal to$

Moderate

A

$(- \frac{1}{e}, \frac{1}{e^{2}})$
B

$(\frac{1}{e}, \frac{1}{e^{2}})$
C

$(\frac{1}{e}, \frac{- 1}{e^{2}})$
D

$(\frac{- 1}{e}, \frac{- 1}{e^{2}})$

Solution

$\begin{aligned} e^{y} + x y = e \Rightarrow e^{y} \frac{d y}{d x} + x \frac{d y}{d x} + y = 0 \\ \frac{d y}{d x} (e^{y} + x) = - y \Rightarrow {(\frac{d y}{d x})}_{(0, 1)} = - \frac{1}{e} \end{aligned}$
Again diff w.r.to ‘x’
$\begin{array}{l} e^{y} \cdot \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} e^{y} \cdot \frac{d y}{d x} + x \frac{d^{2} y}{d x^{2}} + \frac{d y}{d x} + \frac{d y}{d x} = 0 \\ (x + e^{y}) \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} e^{y} + 2 \frac{d y}{d x} = 0 \\ (0, 1) \Rightarrow e \frac{d^{2} y}{d x^{2}} + \frac{1}{e^{2}} e + 2 (- \frac{1}{e}) = 0 \Rightarrow \frac{d^{2} y}{d x^{2}} = \frac{1}{e^{2}} \end{array}$

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