QUIZ #6: CALCULUS 1A (Stankova) Wednesday, March 3, 2004 Section 10:00–11:00 (Voight)

Problem 1. Find y′ if

y = ln (\frac{2x}{x^{2} − y^{2}} ) .

Solution. We have

y′ = \frac{1}{\frac{2x}{x^{2} − y^{2}}} (\frac{2x}{x^{2} − y^{2}} )′

= \frac{x^{2} − y^{2}}{2x} \frac{(x^{2} − y^{2})(2) − 2x(2x − 2yy′)}{(x^{2} − y^{2})^{2}}

= \frac{2x^{2} − 2y^{2} − 4x^{2} + 4xyy′}{2x(x^{2} − y^{2})}

= \frac{−x^{2} − y^{2} + 2xyy′}{x(x^{2} − y^{2})}

x(x^{2} − y^{2})y′ = −x^{2} − y^{2} + 2xyy′
(x^{3} − xy^{2} − 2xy)y′ = −x^{2} − y^{2}
y′ = \frac{−x^{2} − y^{2}}{x^{3} − xy^{2} − 2xy} = \frac{x^{2} + y^{2}}{xy^{2} + 2xy − x^{3}} .

Note the cancellation of the (x^{2} − y^{2}) terms in the third step!

QUIZ #6: CALCULUS 1A (Stankova) Wednesday, March 3, 2004 Section 11:00–12:00 (Voight)

Problem 1. Find an equation of the tangent line to the curve

y = \frac{| ln x|}{x^{2} + 1}

at the point (2, (ln 2)/5). Simplify if you want full credit.

Solution. Since ln 2 > 0, we know that ln x > 0 when x is near x = 2, hence | ln x| = ln x near x = 2. Therefore by the quotient rule, near x = 2 we have

y′ = \frac{(x^{2} + 1)(1/x) − (ln x)(2x)}{(x^{2} + 1)^{2}}

so

y′(2) = \frac{5/2 − 4 ln 2}{25} = \frac{5 − 8 ln 2}{50}.

Therefore by the point-slope formula, the tangent line is

y − \frac{ln 2}{5} = (\frac{5 − 8 ln 2}{50}) (x − 2).

This is enough simplification, really, but we can also write

50y − 10 ln 2 = (5 − 8 ln 2)(x − 2) = (5 − 8 ln 2)x − 10 + 16 ln 2

so

50y = (−8 ln 2 + 5)x + (26 ln 2 − 10).