# Math Help - Quotient rule

1. ## Quotient rule

I need help.

I want to use the quotient rule to find $\frac{d}{dx}[\frac{3t^1^6-4}{t^6}]$

So I do
$\frac{t^6(\frac{d}{dx}[3t^1^6-4]) - (3t^1^6-4)(\frac{d}{dx}[t^6])}{(t^6)^2}$

= $\frac{t^6(48t^1^5) - (3t^1^6-4)(6t^5)}{t^1^2}$
= $\frac{48t^2^1-18t^2^1-24t^5}{t^1^2}$
= $\frac{30^2^1-24t^5}{t^1^2}$

= $30^9-24t^-^7$

but the answer in the book is
= $30^9+24t^-^7$

What Am I doing wrong?

2. ## Re: Quotient rule

Originally Posted by delgeezee
I need help.

I want to use the quotient rule to find $\frac{d}{dx}[\frac{3t^1^6-4}{t^6}]$

So I do
$\frac{t^6(\frac{d}{dx}[3t^1^6-4]) - (3t^1^6-4)(\frac{d}{dx}[t^6])}{(t^6)^2}$

= $\frac{t^6(48t^1^5) - (3t^1^6-4)(6t^5)}{t^1^2}$
= $\frac{48t^2^1-18t^2^1-24t^5}{t^1^2}$
= $\frac{30^2^1-24t^5}{t^1^2}$

= $30^9-24t^-^7$

but the answer in the book is
= $30^9+24t^-^7$

What Am I doing wrong?
Do you have to use the quotient rule? It is far easier to say that
$f(t) = \dfrac{3t^{16}-4}{t^6} = \dfrac{3t^{16}}{t^6} - \dfrac{4}{t^6} = 3t^{10} - 4t^{-6}$

Now you can use the power rule to find $f'(t)$

Your mistake is when you multiplied out the second term in the numerator: $-(3t^{16}-4)(6t^5) = =18t^{21} + 24t^5$

The minus sign becomes a plus because you're subtracting a negative -(-24t^5)

3. ## Re: Quotient rule

I would like to be able to find the solution both ways so that I can double check my finals answers. It gives me a degree of self-assurance. But the alternative way you showed is indeed easier.
Originally Posted by e^(i*pi)
Do you have to use the quotient rule? It is far easier to say that
$f(t) = \dfrac{3t^{16}-4}{t^6} = \dfrac{3t^{16}}{t^6} - \dfrac{4}{t^6} = 3t^{10} - 4t^{-6}$

Now you can use the power rule to find $f'(t)$

Your mistake is when you multiplied out the second term in the numerator: $-(3t^{16}-4)(6t^5) = =18t^{21} + 24t^5$

The minus sign becomes a plus because you're subtracting a negative -(-24t^5)

4. ## Re: Quotient rule

Originally Posted by delgeezee
I need help.

I want to use the quotient rule to find $\frac{d}{dx}[\frac{3t^1^6-4}{t^6}]$

So I do
$\frac{t^6(\frac{d}{dx}[3t^1^6-4]) - (3t^1^6-4)(\frac{d}{dx}[t^6])}{(t^6)^2}$

= $\frac{t^6(48t^1^5) - (3t^1^6-4)(6t^5)}{t^1^2}$
No problem.

= $\frac{48t^{21}-(18t^{21}-24t^5)}{t^{12}}$
Here is where you went wrong, I have put brackets round the second term in the numerator to make it easier to see. What you did is say that $-(18t^{21}-24t^5) = -18t^{21} - 24t^5$ but the minus sign needs to be applied to both terms. I find it helps to imagine the minus sign as -1 and to then use the distributive law:

$-1(18t^{21} - 24t^5) = (-1 \cdot 18t^{21}) + (-1 \cdot -24t^5) = -18t^{21} + 24t^5$

I may have explained this poorly so please post back if you don't understand (and it's good to check both ways)