QUIZ #5: CALCULUS 1A (Stankova) Wednesday, February 25, 2004 Section 10:00–11:00 (Voight)

Problem 1. Use logarithmic differentiation to find the derivative of the function

y = x^{ln x}.

Solution. Taking ln of both sides, we obtain

ln y = ln(x^{ln x}) = ln(x) ln(x) = ln(x)^{2}.

Differentiating now with respect to x, we have (implicit differentiation)

\frac{d}{dx}(ln y) = \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(ln x)^{2}.

To differentiate the right-side, we apply the chain rule:

\frac{d}{dx}(ln x)^{2} = 2 ln x\frac{d}{dx}(ln x) = \frac{2 ln x}{x}.

Therefore we get

\frac{1}{y} \frac{dy}{dx} = \frac{2 ln x}{x}

so

\frac{dy}{dx} = y \frac{2 ln x}{x} = x^{ln x} \frac{2 ln x}{x}.

QUIZ #5: CALCULUS 1A (Stankova) Wednesday, February 25, 2004 Section 11:00–12:00 (Voight)

Problem 1. Differentiate the function

f(x) = sin(ln(2x)).

Solution. We use the chain rule:

f′(x) = cos(ln(2x))(ln 2x)′ = cos(ln(2x))\frac{2}{2x} = \frac{cos(ln(2x))}{x}.

Problem 2. Differentiate the function

h(x) = ln(x + \sqrt{x^{2} − 1}).

Solution. We find the derivative again using the chain rule

h′(x) = (ln(x + \sqrt{x^{2} − 1}))′ = \frac{1}{x + \sqrt{x^{2} − 1}}(x + \sqrt{x^{2} − 1})′.

The second part is

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

= 1 + \frac{x}{\sqrt{x^{2} − 1}} = \frac{x + \sqrt{x^{2} − 1}}{\sqrt{x^{2} − 1}}.

Altogether, this gives the derivative

h′(x) = \frac{1}{x + \sqrt{x^{2} − 1}} · \frac{x + \sqrt{x^{2} − 1}}{\sqrt{x^{2} − 1}} = \frac{1}{\sqrt{x^{2} − 1}}.