QUIZ #4: CALCULUS 1A (Stankova) Wednesday, February 18, 2004 Section 10:00–11:00 (Voight)

Problem 1. Find the derivative of the function

F (x) = \frac{x^{2} + 6\sqrt{x} + 5}{\sqrt{x}}

in two ways: by using the Quotient Rule and by simplifying first. Show that your answers are equivalent.

Solution. By the Quotient Rule, we have

F′(x) = \frac{ho dhi − hi dho}{ho^{2}}

= \frac{\sqrt{x} (2x + \frac{3}{\sqrt{x}}) − (x^{2} + 6\sqrt{x} + 5) (\frac{1}{2\sqrt{x}})}{x}

= \frac{1}{x} (2x\sqrt{x} + 3 − \frac{1}{2}x\sqrt{x} − 3 − \frac{5}{2\sqrt{x}})

= \frac{1}{x} (\frac{3}{2}x\sqrt{x} − \frac{5}{2\sqrt{x}}) = \frac{3}{2\sqrt{x}} − \frac{5}{2x\sqrt{x}}.

Simplifying, we have

F (x) = x^{3/2} + 6 + 5x^{−1/2}

so by the Power Rule,

F′(x) = \frac{3}{2}x^{1/2} − \frac{5}{2}x^{−3/2}.

These are the same, since x^{1/2} = \sqrt{x} and x^{−3/2} = \frac{1}{x\sqrt{x}}.

QUIZ #4: CALCULUS 1A (Stankova) Wednesday, February 18, 2004 Section 11:00–12:00 (Voight)

Problem 1. Find the derivative of the function

f(x) = \frac{2}{x^{2} − x}

using the definition of the derivative. State the domain of the function and the domain of the derivative.

You may not use only differentiation laws!

Solution. By definition,

f′(x) = lim_{h→0} \frac{f(x + h) − f(x)}{h}

= lim_{h→0} \frac{\frac{2}{(x + h)^{2} − (x + h)} − \frac{2}{x^{2} − x}}{h} .

No need to simplify the denominators! We look for the common denominator (the product of both) and combine; the first factor is missing an x^{2} −x and the second is missing a (x + h)^{2} − (x + h), so we get

f′(x) = lim_{h→0} \frac{2(x^{2} − x) − 2((x + h)^{2} − (x + h))}{h(x^{2} − x)((x + h)^{2} − (x + h))}.

Remember, the h enters the denominator since we are dividing by it. Now we multiply out the numerator and hope to cancel the h in the denominator. The numerator is

(2x^{2} − 2x) − 2(x^{2} + 2xh + h^{2} − x − h) = −4xh − 2h^{2} + 2h = h(−4x − 2h + 2),

so cancelling the h, we get

f′(x) = lim_{h→0} \frac{−4x − 2h + 2}{h(x^{2} − x)((x + h)^{2} − (x + h))} = \frac{−4x + 2}{(x^{2} − x)(x^{2} − x)}.

The domain of the function is wherever the denominator does not vanish, x^{2} − x ≠ 0, i.e. x ≠ 0, 1. The same is true of the derivative: its domain is x ≠ 0, 1.

Check your work using the quotient rule:

f′(x) = \frac{(x^{2} − x)(0) − 2(2x − 1)}{(x^{2} − x)^{2}} = \frac{−4x + 2}{(x^{2} − x)^{2}} .

Yup!

The algebra is also not so bad if you use the alternative definition:

f′(a) = lim_{x→a} \frac{f(x) − f(a)}{x − a}

= lim_{x→a} \frac{\frac{2}{x^{2} − x} − \frac{2}{a^{2} − a}}{x − a}

= lim_{x→a} \frac{2(a^{2} − a) − 2(x^{2} − x)}{(x − a)(x^{2} − x)(a^{2} − a)}

= lim_{x→a} \frac{2a^{2} − 2x^{2} + 2x − 2a}{(x − a)(x^{2} − x)(a^{2} − a)}

= lim_{x→a} \frac{2(a − x)(a + x) + 2(x − a)}{(x − a)(x^{2} − x)(a^{2} − a)}

= lim_{x→a} \frac{−2(x + a) + 2}{(x^{2} − x)(a^{2} − a)} = \frac{−4a + 2}{(a^{2} − a)^{2}} .