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Chapter 16: Problem 7

Find \(\frac{\mathrm{d} y}{\mathrm{~d} x}\) given (a) \(x^{2}+x^{3}+y^{2}-y^{3}=1\) (b) \(2 x^{2}-y^{2}+3 x y-7 x-10 y=0\) (c) \(x y^{2}+\frac{y}{x}=\mathrm{e}^{x}\) (d) \(\ln (x y)-\sqrt{x}=\sqrt{y}\) (e) \(x \sin y+y^{2} \cos 2 x=y\)

Short Answer

Expert verified
Question: Find the derivatives of the following implicit functions. a) \(x^2 + x^3 + y^2 - y^3 = 1\) b) \(2x^2 - y^2 + 3xy - 7x - 10y = 0\) c) \(xy^2 + \frac{y}{x} = e^x\) d) \(\ln(xy) - \sqrt{x} = \sqrt{y}\) e) \(x \sin y + y^2 \cos 2x = y\) Answer: a) \(\frac{dy}{dx} = \frac{-2x - 3x^2}{2y - 3y^2}\) b) \(\frac{dy}{dx} = \frac{4x - 7}{3x - 2y - 10}\) c) \(\frac{dy}{dx} = \frac{e^x - y^2 + \frac{y}{x^2}}{2x}\) d) \(\frac{dy}{dx} = \frac{2\sqrt{x} - x}{2\sqrt{y} - y}\) e) \(\frac{dy}{dx} = \frac{\sin y - 2y^2\sin 2x}{1 + 2y\cos 2x - x\cos y}\)

Step by step solution

01

Differentiate both sides of the equation with respect to x

Differentiate the equation \(x^2 + x^3 + y^2 - y^3 = 1\) with respect to \(x\).
02

Apply the chain rule to y terms and find the derivative

Applying the chain rule to \(y^2\) and \(y^3\), we get: \(2x + 3x^2 + 2yy' - 3y^2y' = 0\)
03

Solve for y'

Factor out \(y'\) from the equation: \(y'(2y - 3y^2) = -2x - 3x^2\). Thus, \(\frac{dy}{dx} = y' = \frac{-2x - 3x^2}{2y - 3y^2}\). #For (b)#
04

Differentiate both sides of the equation with respect to x

Differentiate the equation \(2x^2 - y^2 + 3xy - 7x - 10y = 0\) with respect to \(x\).
05

Apply the chain rule to y terms and find the derivative

Applying the chain rule to \(y^2\) and \(y\), we get: \(4x - 2yy' + 3xy' - 7 - 10y' = 0\)
06

Solve for y'

Solve for \(y'\): \(y'(3x - 2y - 10) = 4x - 7\). Thus, \(\frac{dy}{dx} = y' = \frac{4x - 7}{3x - 2y - 10}\). #For (c)#
07

Differentiate both sides of the equation with respect to x

Differentiate the equation \(xy^2 + \frac{y}{x} = e^x\) with respect to \(x\).
08

Apply the chain rule to y terms and find the derivative

Applying the chain rule to \(y^2\) and \(y\), we get: \((y^2 + x\cdot(2yy')) - \frac{y}{x^2} = e^x\)
09

Solve for y'

Solve for \(y'\): \(y'(2x) = e^x - y^2 + \frac{y}{x^2}\). Thus, \(\frac{dy}{dx} = y' = \frac{e^x - y^2 + \frac{y}{x^2}}{2x}\). #For (d)#
10

Differentiate both sides of the equation with respect to x

Differentiate the equation \(\ln(xy) - \sqrt{x} = \sqrt{y}\) with respect to \(x\).
11

Apply the chain rule to y terms and find the derivative

Applying the chain rule to \(y\) and \(\sqrt{y}\), we get: \(\frac{yx + xy'}{xy} - \frac{1}{2\sqrt{x}} = \frac{y'}{2\sqrt{y}}\)
12

Solve for y'

Solve for \(y'\): \(y' = \frac{2\sqrt{x} - x}{2\sqrt{y} - y}\). Thus, \(\frac{dy}{dx} = y' = \frac{2\sqrt{x} - x}{2\sqrt{y} - y}\). #For (e)#
13

Differentiate both sides of the equation with respect to x

Differentiate the equation \(x \sin y + y^2 \cos 2x = y\) with respect to \(x\).
14

Apply the chain rule to y terms and find the derivative

Applying the chain rule to \(\sin y\), \(y^2\), and \(y\), we get: \(\sin y + x \cos y y' + (-2y^2\sin 2x - 2y \cos 2x y') = y'\)
15

Solve for y'

Solve for \(y'\): \(y'(1 + 2y\cos 2x - x\cos y) = \sin y - 2y^2\sin 2x\). Thus, \(\frac{dy}{dx} = y' = \frac{\sin y - 2y^2\sin 2x}{1 + 2y\cos 2x - x\cos y}\).

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