Find f'(x) if f(x)=ln [(sqrt (x^2+3)] / (x(4x^2-1)^3) find the derivative! show steps/work so i can follow along. you!

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))

- "runner b is moving at a slower rate."

- "the graphs shown are graphs of distance(km) against time(min).

the slopes of this graphs are are given by:

slope = (change in y)/(change in x)

= (change in distance)/(change in time)

this is the formula for finding the speed. so, the slope of these graphs represent the speed of the runner.

taking each division in the graph to represent 1 on both axis:

runner a ⇒ speed = (2-0)/(1-0)

= 2/1

= 2 km/min

runner b ⇒ speed = (3-0)/(3-0)

= 3/3

= 1 km/min

this shows that   runner b is moving at a slower rate."