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CHAP. III.
ALGEBRAIC FUNCTIONS
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Using V, with u = x^2 + 1 , v = x^2 - 1 ,
\begin{aligned}dy &= \frac{(x^2 - 1) d(x^2 + 1) - (x^2 + 1) d(x^2 - 1)}{(x^2 - 1)^2} \\&= \frac{(x^2 - 1) 2x dx - (x^2 + 1) 2x dx}{(x^2 - 1)^2} \\&= -\frac{4x dx}{(x^2 - 1)^2}.\end{aligned}
\text{Ex. 5. } y = \sqrt{x^2 - 1}.
Using VI, with u = x^2 - 1 ,
\begin{aligned}\frac{dy}{dx} &= \frac{d}{dx} (x^2 - 1)^{\frac{1}{2}} = \frac{1}{2} (x^2 - 1)^{-\frac{1}{2}} \frac{d}{dx} (x^2 - 1) \\&= \frac{1}{2} (x^2 - 1)^{-\frac{1}{2}} (2x) = \frac{x}{\sqrt{x^2 - 1}}.\end{aligned}
\text{Ex. 6. } x^2 + xy - y^2 = 1.
We can consider y a function of x determined by the equation. Then
\begin{aligned}d(x^2) + d(xy) - d(y^2) &= d(1) = 0, \\ \text{that is,} \quad 2x dx + x dy + y dx - 2y dy &= 0, \\ (2x + y) dx + (x - 2y) dy &= 0.\end{aligned}
Consequently,
\frac{dy}{dx} = \frac{2x + y}{2y - x}.
\text{Ex. 7. } x = t + \frac{1}{t}, \quad y = t - \frac{1}{t}.
In this case
dx = dt - \frac{dt}{t^2}, \quad dy = dt + \frac{dt}{t^2}.
\frac{dy}{dx} = \frac{1 + \frac{1}{t^2}}{1 - \frac{1}{t^2}} = \frac{t^2 + 1}{t^2 - 1}.
Ex. 8. Find an approximate value of y = \left(\frac{1-x}{1+x}\right)^{\frac{1}{3}} when x = 0.2 .
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