Question

Find f'(x) if f(x)=ln [(sqrt (x^2+3)] / (x(4x^2-1)^3)

Please help find the derivative! Show steps/work so I can follow along. Thank you!

Answer 1

`f'(x)=(8x^4-2x^2-\left(28x^4+83x^2-3\right)ln(\left(x^2+3\right)))/(2x^2\left(x^2+3\right)\left(4x^2-1\right)^4)`

Step-by-step explanation:

It can work well to consider the function in parts. Define the following:

... a(x) = (1/2)ln(x^2+3)

... b(x) = x(4x^2-1)^3

Then the derivatives of these are ...

... a'(x) = (1/2)·1/(x^2 +3)·2x = x/(x^2+3)

... b'(x) = (4x^2 -1)^3 + 3x(4x^2 -1)^2·8x = (4x^2 -1)^2·(4x^2 -1 +24x^2)

... = (4x^2 -1)^2·(28x^2 -1)

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Putting the parts together

f(x) = a(x)/b(x)

f'(x) = (b(x)a'(x) -a(x)b'(x))/b(x)^2 . . . . . rule for quotient of functions

Substituting values, we have

... f'(x) = (x(4x^2 -1)^3·x/(x^2 +3) -(1/2)ln(x^2 +3)·(4x^2 -1)^2·(28x^2 -1)) / (x(4x^2 -1)^3)^2

We can cancel (4x^2 -1)^2 from numerator and denominator. We can also eliminate fractions (1/2, 1/(x^2+3)). Then we have ...

... f'(x) = 2x^2(4x^2 -1) -(x^2 +3)ln(x^2 +3)·(28x^2 -1)/(2x^2·(x^2 +3)(4x^2 -1)^4))

Simplifying a bit, this becomes ...

... f'(x) = (8x^4 -2x^2 -ln(x^2 +3)·(28x^4 +83x^2 -3))/(2x^2·(x^2 +3)(4x^2 -1)^4))