How do you find the derivative of #y=(2x^-1-x^-2)/(3x^-1-4x^-2)#?

Question

Scarlett Allison · Accepted Answer

Note the domain and rewrite, then use the quotient rule.

#f(x) = y = (2x^-1-x^-2)/(3x^-1-4x^-2)# Domain of #f# is all reals except #0#, and #4/3#. Now use algebra to rewrite #f(x) = ((2x^-1-x^-2)x^2)/((3x^-1-4x^-2)x^2)# # = (2x-1)/(3x-4)# Now use the quotient rule #f'(x) = ((2)(3x-4)-(3)(2x-1))/(3x-4)# # = (-5)/(3x-4)^2# If we want to make sure the domain is clear, then multiply to write #f'(x) = (-5x^-1)/((3x-4)^2x^-1)#

Emilia Aguilar · Answer

undefined To find the derivative of the given function, $ y = \frac{{2x^{-1} - x^{-2}}}{{3x^{-1} - 4x^{-2}}} $, you can use the quotient rule. The quotient rule states that if you have a function $ u(x) $ divided by another function $ v(x) $, then the derivative is given by:

$$ \frac{{d}}{{dx}}\left(\frac{{u(x)}}{{v(x)}}\right) = \frac{{u'(x)v(x) - u(x)v'(x)}}{{[v(x)]^2}} $$

Where $ u'(x) $ and $ v'(x) $ represent the derivatives of $ u(x) $ and $ v(x) $ respectively.

So, applying the quotient rule to the given function:

$$ y' = \frac{{(2x^{-1})'(3x^{-1} - 4x^{-2}) - (2x^{-1} - x^{-2})(3x^{-1} - 4x^{-2})'}}{{(3x^{-1} - 4x^{-2})^2}} $$

$$ y' = \frac{{(-2x^{-2})(3x^{-1} - 4x^{-2}) - (2x^{-1} - x^{-2})(-3x^{-2} + 8x^{-3})}}{{(3x^{-1} - 4x^{-2})^2}} $$

$$ y' = \frac{{-6x^{-3} + 8x^{-4} + 6x^{-3} - 16x^{-4}}}{{(3x^{-1} - 4x^{-2})^2}} $$

$$ y' = \frac{{2x^{-4}}}{{(3x^{-1} - 4x^{-2})^2}} $$