if y = (x - 3)(x^ 2 + 4) 3x^ 2 + 4x + 5 , find dy dx - Vidyadip

Question

$\text{if y } = \sqrt{\frac{(x - 3)(x^{2} + 4)}{3x^{2} + 4x + 5}}, \text{find} \frac{dy}{dx}$

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Answer

$y = \sqrt{\frac{(x - 3)(x^{2} + 4)}{3x^{2} + 4x + 5}} \dots\dots\dots \text{(i})$

Talking log of both sides of (i), we get

$\text{\log y} = \frac{1}{2} \bigg[\log (x - 3) + \log (x^{2} + 4) - \log (3x^{2} + 4x +5)\bigg]$

$\therefore \frac{1}{y} \frac{dy}{dx} = \frac{1}{2}\bigg[ \frac{1}{x-3} + \frac{2x}{x^{2} + 4} - \frac{6x + 4}{3x^{2} + 4x + 5}\bigg]$

$\therefore \frac{dy}{dx} = \frac{y}{2}\bigg[ \frac{1}{x-3} + \frac{2x}{x^{2} + 4} - \frac{6x + 4}{3x^{2} + 4x + 5}\bigg]$

OR

$\frac{1}{2} \sqrt{\frac{x - 3)(x ^{2} + 4)}{3x^{2} + 4x +5}} \bigg[ \frac{1}{x - 3} + \frac{2x}{x^{2} + 4} - \frac{6x + 4}{3x^{2} + 4x + 5}\bigg]$

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