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Q.
Differentiate $ {{x}^{-3}}\,(2+7x) $
J & K CETJ & K CET 2014Limits and Derivatives
Solution:
Let $ y={{x}^{-3}}(2+7x)=2{{x}^{-3}}+7{{x}^{-2}} $ On differentiating w .r .t .x, we get $ \frac{dy}{dx}=2(-3){{x}^{-3-1}}+7(-2){{x}^{-2-1}} $
$=-6{{x}^{-4}}-14{{x}^{-3}} $
$=-2\left( \frac{3}{{{x}^{4}}}+\frac{7}{{{x}^{3}}} \right)=\frac{-2}{{{x}^{4}}}(3+7x) $