Question

Which is the derivative of \[\sqrt{\frac{(x-3)(x^2+4)}{3x^2+4x+5}}\]?

Options

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

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

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

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

Answer 1

Multiply \[\frac{1}{y}\frac{dy}{dx}\] by the original value of \[y\]. The factor \[\frac{1}{2}\] remains because the original function is a square root.