Question

What is \[\frac{1}{y}\cdot\frac{dy}{dx}\] for \[y=\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}\]?

Options

• \[\frac{1}{2}\left[\frac{1}{x-3}-\frac{2x}{x^2+4}-\frac{6x+4}{3x^2+4x+5}\right]\]

• \[\frac{1}{2}\left[\frac{1}{x-3}+\frac{2x}{x^2+4}-\frac{6x+4}{3x^2+4x+5}\right]\]

• \[\frac{1}{2}\left[\frac{1}{x-3}+\frac{x^2}{x^2+4}-\frac{3x^2+4}{3x^2+4x+5}\right]\]

• \[\frac{1}{2}\left[\frac{1}{x-3}+\frac{2x}{x^2+4}+\frac{6x+4}{3x^2+4x+5}\right]\]

Answer 1

Differentiate each logarithm using the chain rule. The denominator logarithm is subtracted, so its derivative \[\frac{6x+4}{3x^2+4x+5}\] has a negative sign.