What is the derivative of this function

f(t) = (4t) / (5 + t^2)

I got (20 - 4t^2) / (25 + t^4) but im wrong.

The derivative of a fraction f(x) = g(x) / h(x) is

f'(x) = [g'(x) * h(x) - g(x) * h'(x)] / (h(x))^2

g(x) = 4t
g'(x) = 4

h(x) = 5 + t^2
h'(x) = 2t

So

f'(x) = [4 * (5 + t^2) - 4t * 2t] / (5 + t^2)^2
= (20 + 4t^2 - 8t^2) / (25 + 10t^2 + t^4)
= (-4t^2 + 20) / (t^4 + 10t^2 + 25)


Now, personally, I always had a problem remembering if it was g'h = gh' or gh' - g'h. So I did these using the product rule, like this.

f'(t) = g'(t) * h(t) + g(t) * h'(t)

g(t) = 4t
g'(t) = 4

h(t) = (5 + t^2)^(-1)
h'(t) = -1 * (5 + t^2)^(-2) * 2t

f'(t) = 4t * [-2t * (5 + t^2)^(-2)] + 4 * (5 + t^2)^(-1)
= [-8t^2 * (5 + t^2)^(-2)] + 4 * (5 + t^2) * (5 + t^2)^(-2)
= (-8t + 20 + 4t) / (5 + t^2)^2

Which as you can see will eventually give us the same answer.

Quotient Rule:

[f'(x)g(x) - f(x)g'(x)]/[(g(x))^2]

So:

[4(5+t^2) - (4t)(2t)]/[(5+t^2)^2]

Simplify:

[20 + 4t^2 - 8t^2]/[25 + 10t^2 + t^4]

More:

[20 - 4t^2]/[25 + 10t^2 + t^4]

That is as simplified as it gets...

Also when simplifying the denominator remember that (5 + t^2)^2 = (5 + t^2)(5 + t^2).

Hope this helps.