How do you find the derivative of #[(2x^3) - (4x^2) + 3] / x^2 #?

1 Answer
Zach J.
Mar 4, 2018

#d/dx({2x^3-4x^2+3}/x^2)={2(x^3-3)}/x^3#

Explanation:

We use the quotient rule, which says for #f(x)=g(x)/{h(x)}, f'(x)={h(x)g'(x)-h'(x)g(x)}/{(h(x))^2}#.

So, we first let #f(x)={2x^3-4x^2+3}/x^2#.

Thus, #g(x)=2x^3-4x^2+3# and #h(x)=x^2#.

Next, we find #g'(x)# and #h'(x)#

This just involves the power rule, so
#g'(x)=3*2x^2-2*4x+0=6x^2-8x#
#h'(x)=2x#

Now, we just plug these values into the quotient rule to get the following:
#f'(x)={(x^2)(6x^2-8x)-(2x)(2x^3-4x^2+3)}/{(x^2)^2}#
#f'(x)={(6x^4-8x^3)-(4x^4-8x^3+6x)}/{x^4}#
#f'(x)={6x^4-4x^4-8x^3+8x^3-6x}/{x^4}#
#f'(x)={2x^4-6x}/{x^4}#
#f'(x)={2x(x^3-3)}/{x^4}#
#f'(x)={2(x^3-3)}/x^3#
Therefore, #d/dx({2x^3-4x^2+3}/x^2)={2(x^3-3)}/x^3#

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