Differentiate $e^{9x}/(-8x^5-4x)$

Question

I have used the quotient rule to differentiate the expression, but I'm not sure how to simplify from here:

$$\frac{ (9e^{9x})*(-8x^5-4x) - (e^{9x}*-40x^4-4) }{(-8x^5-4x)^2}$$

Answer 1

$$\frac{ (9e^{9x})*(-8x^5-4x) - (e^{9x}*-40x^4-4) }{(-8x^5-4x)^2}$$ Pay attention to your parentheses $$\frac{ (9e^{9x})*(-8x^5-4x) - e^{9x}*(-40x^4-4) }{(-8x^5-4x)^2}$$ $$\frac{ (9e^{9x})*(-2x^5-x) - e^{9x}*(-10x^4-1) }{(4x^5+2x)^2}$$ $$\frac{ e^{9x}(-18x^5+10x^4-9x+1) }{4x^2(2x^4+1)^2}$$

Answer 2

After getting to the following result by doing some math:

$$\frac{ e^{9x}(-18x^5+10x^4-9x+1) }{(4x^5+2x)^2}= \frac{ e^{9x}(-18x^5+10x^4-9x+1)} {16x^{10}+16x^6+4x^2}$$

you can perform long division of the polynomial of the denominator by the polynomial of the denominator. And this is a much more simplified from.