Quotient Rule Problem - # 3

[Jason76]

$\displaystyle f(t) = \dfrac{t}{(t - 2)^{2}}$

$\displaystyle f'(x) = [(t - 2)^{2}][1] - [t][2(u) du]$

$\displaystyle f'(x) = [(t - 2)^{2}][1] - [t][2(u)(1)]$

$\displaystyle f'(x) = [(t - 2)^{2}][1] - [t][2(u)]$

$\displaystyle f'(x) = [(t - 2)^{2}][1] - [t][2(t - 2)]$

$\displaystyle f'(x) = [(t - 2)^{2}][1] - [t][2t - 4)]$


$\displaystyle f'(x) = (t - 2)^{2}(1) - 2t^{2} - 4$

$\displaystyle f'(x) = (t - 2)^{2} - 2t^{2} - 4$

 

[Deleted member 4993]

$\displaystyle f(t) = \dfrac{t}{(t - 2)^{2}}$

$\displaystyle f'(x) = [(t - 2)^{2}][1] - [t][2(u) du]$ ..........What is 'u'? please define it

$\displaystyle f'(x) = [(t - 2)^{2}][1] - [t][2(u)(1)]$

$\displaystyle f'(x) = [(t - 2)^{2}][1] - [t][2(u)]$

$\displaystyle f'(x) = [(t - 2)^{2}][1] - [t][2(t - 2)]$ ..........Whatever you are doing - this is incorrect

$\displaystyle f(t) = \dfrac{t}{(t - 2)^{2}}$

$\displaystyle f(t) \ = \ t * (t - 2)^{-2}$

$\displaystyle \dfrac{d[f(t)]}{dt} \ = \ 1 * (t - 2)^{-2} + (-2)(t-2)^{-3}$

$\displaystyle \dfrac{d[f(t)]}{dt} \ = \ (t - 2)^{-3} * [t - 2 -2]$

$\displaystyle \dfrac{d[f(t)]}{dt} \ = \ (t - 2)^{-3} * [t - 4]$

 

[HallsofIvy]

You both seem to be under the impression that (f/g)'= f'g- fg'.

It isn't- the quotient rule says that (f/g)'= (f'g- fg')/g^2. You have forgotten the denominator, g^2.

 

[Deleted member 4993]

You both seem to be under the impression that (f/g)'= f'g- fg'.

It isn't- the quotient rule says that (f/g)'= (f'g- fg')/g^2. You have forgotten the denominator, g^2.

No - I converted (f/g) to f * (g)^-1

And then applied the product rule

d/dt [f/g] = d/dt [f * (g)^-1] = f' * (g)^-1 + f * (-1) * (g)^-2 * g'

which can be converted to:

f' * (g)^-1 + f * (-1) * (g)^-2 * g'

= f' / (g) - f * g'/(g)²

$\displaystyle = \dfrac{f' * g - f * g'}{g^2}$

I remember the product rule only - I derive the quotient rule when necessary.

 

[Jason76]

I don't think the quotient rule is hard to memorize. It's the same thing as the product rule (assuming (with regards to the product rule) that you start on the left with $\displaystyle g$ instead of $\displaystyle f$). The only difference (with regards to the quotient rule) is the minus sign in the middle, and the $\displaystyle g^{2}$ on the bottom.

$\displaystyle g(f') + f(g')$ One of the two ways to write the product rule.

$\displaystyle \dfrac{g(f') - f(g')}{g^{2}}$ - the quotient rule (very similar with two differences)

 

[Deleted member 4993]

I don't think the quotient rule is hard to memorize. It's the same thing as the product rule (assuming (with regards to the product rule) that you start on the left with $\displaystyle g$ instead of $\displaystyle f$). The only difference (with regards to the quotient rule) is the minus sign in the middle, and the $\displaystyle g^{2}$ on the bottom.

$\displaystyle g(f') + f(g')$ One of the two ways to write the product rule.

$\displaystyle \dfrac{g(f') - f(g')}{g^{2}}$ - the quotient rule (very similar with two differences)

You can follow any procedure you want to - as long as it provides correct answer and is applied without violating fundamental rules of mathematics.

However, in the original post of this thread, you did not apply the quotient rule correctly - thus you got incorrect answer.