Asked by Eileen
Differentiate the following function
(1/x)-(2/x^2)+(3/x^3)
I got (-9x-3)/x^5
(1/x)-(2/x^2)+(3/x^3)
I got (-9x-3)/x^5
Answers
Answered by
bobpursley
You should be able to do this in your head, if not, something is wrong, all there are the power formula.
f'=-x<sup>-2</sup>+4x<sup>-3</sup>-9x<sup>-4</sup>
check that.
f'=-x<sup>-2</sup>+4x<sup>-3</sup>-9x<sup>-4</sup>
check that.
Answered by
Reiny
I don't see how you got that ...
let y = (1/x)-(2/x^2)+(3/x^3)
= x^-1 - 2x^-2 + 3x^-3
dy/dx = - x^-2 + 4x^-3 - 9x^-4
= -1/x^2 + 4/x^3 - 9/x^4
= (-x^2 + 4x - 9)/x^4 , if you want a common denominator
let y = (1/x)-(2/x^2)+(3/x^3)
= x^-1 - 2x^-2 + 3x^-3
dy/dx = - x^-2 + 4x^-3 - 9x^-4
= -1/x^2 + 4/x^3 - 9/x^4
= (-x^2 + 4x - 9)/x^4 , if you want a common denominator
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