I am given the equation: $x^{a}y^{b} = 6$

Using implicit differentiation I find that the derivative of the equation with respect to y gives: $\frac{d}{dx}(y) = -\frac{ay}{bx}$. However when I attempt to differentiate the "regular" way I don`t seem to reach the same answer. I would greatly appreciate someone walking me through the problem.

  • $\begingroup$ is here assumed to be $$y=y(x)?$$ $\endgroup$ – Dr. Sonnhard Graubner Oct 3 '17 at 20:05
  • 2
    $\begingroup$ $x^ay^b$ is an expression not an equation $\endgroup$ – Riley Oct 3 '17 at 20:05
  • $\begingroup$ You are right, I edited it now. Yes, y =y(x) $\endgroup$ – user487324 Oct 3 '17 at 20:09
  • $\begingroup$ now it is an equation $\endgroup$ – Dr. Sonnhard Graubner Oct 3 '17 at 20:09

then we get $$ax^{a-1}y^b+x^aby^{b-1}y'=0$$ you must use the product and chain rule: $$(uv)'=u'v+uv'$$ and since $$y=y(x)$$ the first derivative is given by $$y'(x)$$ $$(x^a)'=ax^{a-1}(x)$$ and $$(y(x)^b)'=by(x)^{b-1}\cdot y'(x)$$

  • $\begingroup$ Could you differentiate it regularly for me? Preferably step-by-step. $\endgroup$ – user487324 Oct 3 '17 at 20:13
  • $\begingroup$ Thank you for the help! $\endgroup$ – user487324 Oct 3 '17 at 20:21
  • $\begingroup$ it is nice that i could help you! all the best for you! $\endgroup$ – Dr. Sonnhard Graubner Oct 3 '17 at 20:23

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