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how differentiate this question (x^2-4)(x^2+3)^6
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use product rule \[\frac{d}{dx}(uv)=v \frac{du}{dx}+u \frac{dv}{dx}\]
\[\frac{d}{dx}(x^2-4)(x^2+3)^6=(x^2+3)^6 \frac{d}{dx}(x^2-4)+(x^2-4) \frac{d}{dx}(x^2+3)^6\]
= 2x (x²+3)^6 + 12x (x²-4) ( x² + 3) ^5 = 2x ( x² + 3) ^5 * [ x² + 3) + 6 (x²-4) ] = 2x ( x² + 3) ^5 * ( 7x² - 21)
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