Differentiate this f(t)= 2t/(4+t^2)?

Question

Differentiate this f(t)= 2t/(4+t^2)

do we use product rule and quotient rule or how???

Answer 1

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

Since we have a quotient (division) of functions here, we have to use the quotient rule.

In words, the quotient rule goes as follows: "The derivative of the top times the bottom minus the derivative of the bottom times the top, over the bottom _squared_."

If you wish, I'll use square brackets to denote the functions I differentiate, and round brackets for the functions I don't differentiate.

f'(t) = { [2] (4 + t^2) - [2t] (2t) } / (4 + t^2)^2

I'll leave this up to you to simplify. On an exam, they would normally not ask you to simplify.

Answer 2

Use the quotient rule: d(f(x)/g(x)) = (g(x)*f'(x) - f(x)*g'(x)) / (g(x)^2)

Let f(t) =2t
Let g(t) = 4 + t^2

Then f'(t) = 2 and g'(t) = 2t

Plugging the functions into the quotient rule formula,

((4 + t^2)*(2) - (2t)*(2t) ) / (4 + t^2)^2

This gives: (8 + 2t^2 - 4t^2) / (16 + 8t^2 + t^4) Now combine the like terms. The end result should be:

d(f(t)/g(t) = (8 - 2t^2) / (16 + 8t^2 + t^4)

Note that the answer doesn't depend on your choice of which expression is f(x) and which one is g(x). They will yield the same answer, but sometimes one choice is easier to compute than the other.

Answer 3

We can just use Quotient Rule for this one

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

f'(t) = (((4 + t^2)(2t)') - ((4 + t^2)'(2t)))/((4 + t^2)^2)
f'(t) = (2(t^2 + 4) - 2t(2t))/((t^2 + 4)^2)
f'(t) = (2t^2 + 8 - 4t^2)/((t^2 + 4)^2)
f'(t) = (-2t^2 + 8)/((t^2 + 4)^2)
f'(t) = -2(t^2 - 4)/((t^2 + 4)^2)

since the problem can't be factored

f'(t) = (-2t^2 + 8)/((t^2 + 4)^2) or (-2t^2 + 8)/(t^4 + 8t^2 + 16)

but i prefer leaving it as f'(t) = (-2t^2 + 8)/((t^2 + 4)^2)

Answer 4

you can use the product or quotient rule

PRODUCT RULE: you would have to inverse the denominator
f(t)= 2t / (4 + t^2)
f(t) = 2t * (4 + t^2)^-1
f '(t) = [2t * - (4 + t^2)^-2 * 2t] + [(4 + t^2)^-1 * 2 ]
I don't feel like simplifying

QUOTIENT RULE
g' * h - g * h' / h^2

f '(t) = ([(4 + t^2) * 2] - [2t * (2^t)]) / (4 + t^2)^2
I don't feel like simplifying

Answer 5

slightly DiFFERENT METHOd

numerator is the differential of denominator
d/dt(4+t^2) = 2t

so quotient is of the form = f' / f
quotient rule d/dt(f' /f) = f".f - f'.f' / f^2
= f"/f- (f'/f)^2
f= 4+ x^2 : f' = 2x :f" = 2

substitute to get result