How do I find the f'(x) and f''(x) for the given function? 2x? + ху? %3D 1

Name: How do I find the f'(x) and f''(x) for the given function? 2x? + ху? %
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Description: How do I find the f'(x) and f''(x) for the given function? 2x? + ху? %3D 1

Answered: 2x? + ху? %3D 1

Algebra: Structure And Method, Book 1

(REV)00th Edition

ISBN:9780395977224

Author:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Publisher:Richard G. Brown, Mary P. Dolciani, Robert H. Sorgenfrey, William L. Cole

Chapter3: Solving Equation And Problems

Section3.3: Using Several Transformations

Problem 19WE

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How do I find the f'(x) and f''(x) for the given function?

Expert Solution

Step 1

Given function is $2 x^{2} + x y^{2} = 1$ .

BY product rule of differentiation, we have

$\frac{d}{d x} (u v) = u \frac{d}{d x} (v) + v \frac{d}{d x} (u)$

Differentiating the given function with respect to $x$ ,

$\begin{array}{rcl} \frac{d}{d x} (2 x^{2} + x y^{2}) & = & \frac{d}{d x} (1) \\ \frac{d}{d x} (2 x^{2}) + \frac{d}{d x} (x y^{2}) & = & 0 \\ 4 x + (x \frac{d}{d x} (y^{2}) + y^{2} \frac{d}{d x} (x)) & = & 0 \\ 4 x + (x \cdot 2 y \frac{d y}{d x} + y^{2} \cdot 1) & = & 0 \\ 2 x y \frac{d y}{d x} + y^{2} & = & - 4 x \\ 2 x y \frac{d y}{d x} & = & - 4 x - y^{2} \\ \frac{d y}{d x} & = & \frac{- 4 x - y^{2}}{2 x y} \\ y' & = & - \frac{4 x}{2 x y} - \frac{y^{2}}{2 x y} \\ y' & = & - \frac{2}{y} - \frac{y}{2 x} \end{array}$

Therefore, $f' (x) = - \frac{2}{y} - \frac{y}{2 x}$

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