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If y = uv , where u and v are functions of x,
then
\( \normalsize \frac{dy}{dx} = \scriptsize u \normalsize \frac{dv}{dx} + \scriptsize v \normalsize \frac{du}{dx}\)
Example: Differentiate \( \scriptsize (3x -2)(x^2 + 3) \) with respect to x
let u = (3x – 2) and v = x2 + 3
\( \frac{du}{dx} \scriptsize = 3\) \( \frac{dv}{dx} \scriptsize = 2x \) \( \normalsize \frac{dy}{dx} = \scriptsize u \normalsize \frac{dv}{dx} \times \scriptsize v \normalsize \frac{du}{dx}\) \( \frac{dy}{dx} = \scriptsize (3x \; – \; 2) (2x) + (x^2 + 3) (3)\) \( \frac{dy}{dx} = \scriptsize 6x^2 -4x + 3x^2 + 9 \)=\( \frac{dy}{dx} = \scriptsize 9x^2 – 4x + 9 \)
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