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Finding the derivative?

0 votes
f(x)=1/x^6-1/x^8 using negative powers of x.
I know the answer is f '(x)=-6/x^2+8/x^9 but how do I get it?
asked May 11, 2013 in CALCULUS by andrew Scholar

2 Answers

0 votes

1) Given f(x) = 1/x^6 - 1/x^8

Rewrite the given expression in negative powers of x

f(x) = x^-6 - x^-8

Differentiating with respect to x

f'(x) = d/dx(x^-6 - x^-8)

f'(x) = d/dx(x^-6) - d/dx(x^-8)

Recall the formula of differentiation : d/dx(x^n) = n*x^n-1

f'(x) = -6*x^-6-1 - ( - 8*x^-8-1)

f'(x) = -6*x^-7 - ( -8*x-9 )

f'(x) = -6*x^-7 + 8*x^-9

Rewriting the expression inpositive powers of x

f'(x) = -6/x^7 + 8/x^9

answered May 13, 2013 by jeevitha Novice
0 votes

Given f(x) = 1 / x6 - 1/ x8

Rewrite the given expression in negative powers of x

f(x) = x-6- x-8

Diferenciate with respective x

f ' (x) = d / dx(x-6) - d / dx(x-8)

Recall : Differciation formula d / dx (xn) = nxn-1

f ' (x) = (-6x-6-1) - ( -8x-8-1)

         = (-6x-7) - (-8x-9)

         = (-6 / x7) - (-8 / x9)

         = (-6 / x7) + (8 / x9)

Given answer is wrong

f ' (x) = (-6 / x7) + (8 / x9).

 

 

answered May 13, 2013 by diane Scholar

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