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I am doing a problem where I am differentiating from first principles, but I can't simplify the final expression:

$\frac{-2xh - h^2}{x^4h + 2x^3h^2+x^2h^3}$

Could someone explain it in steps?

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It is $$\frac{-2x-h}{x^2(x^2+2xh+h^2)}=\frac{-2x-h}{x^2(x+h)^2}$$ it must be $$x\ne 0,x\ne -h$$ If $$h$$ tends to zero we obtain $$\frac{-2x}{x^2(x^2)}=\frac{-2x}{x^4}=\frac{-2}{x^3}$$

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  • $\begingroup$ Could you explain what would happen to the equation if h tends to 0 please? $\endgroup$
    – Newbie101
    May 29 '19 at 10:46
  • $\begingroup$ Ohh I understand it. Thank you :) $\endgroup$
    – Newbie101
    May 29 '19 at 10:50
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Factorising h from both N&D to get$$=-\frac{h+2x}{x^2(x^2+2xh+h^2)}$$ $$=-\frac{h+2x}{x^2(x+h)^2}$$

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    $\begingroup$ But you factorised $x^2$ on the bottom? Also, when I factorised h from the top I got $h(-2x-h)$ $\endgroup$
    – Newbie101
    May 29 '19 at 10:36
  • $\begingroup$ after cancelling h we can further take x^2 common... in the denoimnator $\endgroup$ May 29 '19 at 10:38
  • $\begingroup$ You can also factorize the denominator $$h(x^4+2x^3h+x^2h^2)$$ $\endgroup$ May 29 '19 at 10:39
  • $\begingroup$ i took minus sign common from the numerator... it cna be done in a single step $\endgroup$ May 29 '19 at 10:39
  • $\begingroup$ @Newbie101 They're the same, $-2x-h = -(2x+h) = -(h+2x)$ $\endgroup$
    – Ak.
    May 29 '19 at 10:43

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