How do I solve for the derivative of 1/(4x^4-6x^2-3)?

Question

i get the same answer as the first response, but when i submit it into the program, it doesnt say it is correct

Answer 1

That is, (4x^4-6x^2-3)^(-1). To solve this, you bring the -1 in front as the coefficient, and then -2 becomes the power that you have raised the inside function to. Then, you take the derivative of the inside. So, (-1)(4x^4-6x^2-3)^(-2) * (16x^3-12x), or -(16x^3-12x)/(4x^4-6x^2-3)^(2). If anything can be factored, factor it.

Answer 2

Switch the bottom to the top and negate the power, so

(4x^-4) - (6x^-2) - 1/3


and then just take a derivative like normal

so

(-16x^-5) + (12x^-3). The 1/3 gets left out because there is no x

Then just flip the equation back, so you have positive powers:

1/ (-16x^5 + 12x^3)

Answer 3

First since its a fraction, undo the fraction part (take it's inverse). So it becomes 4x^(-4) - 6x^(-2) - (1/3). Now just multiply exponent by coefficient and "bump the power down" so it should now be -16x^(-5)+12x^(-3) (the number goes away).