# Thread: Help with the qoutient rule.

1. ## Help with the qoutient rule.

Find dy/dx where y = (x^2 - 5)/(cos(x))

I used the quotient rule and got (Cos x)(2x)-(x^2-5)(-sinx)/(cosx)^2 = (2cosx^2 +sinx^3 + 5 sin x)/ ((cosx)^2)

Is it possible to break this down further algebraically?

2. Originally Posted by Murphie
Find dy/dx where y = (x^2 - 5)/(cos(x))

I used the quotient rule and got (Cos x)(2x)-(x^2-5)(-sinx)/(cosx)^2 = (2cosx^2 +sinx^3 + 5 sin x)/ ((cosx)^2)

Is it possible to break this down further algebraically?
You could do it this way if you wanted to, avoiding the quotient rule altogether. From Wolfram Alpha

3. Originally Posted by Murphie
Find dy/dx where y = (x^2 - 5)/(cos(x))

I used the quotient rule and got (Cos x)(2x)-(x^2-5)(-sinx)/(cosx)^2 = (2cosx^2 +sinx^3 + 5 sin x)/ ((cosx)^2)

Is it possible to break this down further algebraically?
Algebraically this can be reduced by some standard trigonomic identities. So we have:

$\frac{2cos(x^2)+sin(x^3)+5sin(x)}{cos(x)cos(x)}$

By splitting up the fractions we get the following:

$\frac{2cos(x^2)}{cos(x)cos(x)} + \frac{sin(x^3)}{cos(x)cos(x)} + \frac{5sin(x)}{cos(x)cos(x)}$

the last term in the above expression reduces to:

$5sec(x)tan(x)$

Other then that I'm not sure how much more reducing can be done, but the post above mine gives a different technique for going about the differentiation that gives a simpler seeming answer, so thats an option too.