Help with the qoutient rule.

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?

undefined

MHF Hall of Honor
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

Attachments

• 22.8 KB Views: 52

mfetch22

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:

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

By splitting up the fractions we get the following:

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

the last term in the above expression reduces to:

$$\displaystyle 5sec(x)tan(x)$$

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.

Similar threads