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Home  >>  CBSE XI  >>  Math  >>  Limits and Derivatives
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Find the derivative of $x^{-4}(3-4x^{-5})$

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By Leibnitz product rule,
$f'(x) =x^{-4} \large\frac{d}{dx}$$(3-4x^{-5})+(3-4x^{-5})\large\frac{d}{dx}$$(x^{-4})$
$ = x^{-4} \{0 - 4 -(-5)x^{-5-1} \} + (3-4x^{-5})(-4)x^{-4-1}$
$ = x^{-4}(20x^{-6})+(3-4x^{-5})(-4x^{-5})$
$ = 20x^{-10}-12x^{-5}+16x^{-10}$
$ = 36x^{-10}-12x^{-5}$
$ = -\large\frac{12}{x^5}$$+ \large\frac{36}{x^{10}}$
answered Apr 9, 2014 by thanvigandhi_1
 
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