How would you find the derivative of (5cos(x))/(4-sec(x))?

Thanks.

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- Jan 10th 2010, 02:37 PMsuchgreatheightsDerivative Involving Chain Rule
How would you find the derivative of (5cos(x))/(4-sec(x))?

Thanks. - Jan 10th 2010, 02:43 PMJhevon
chain rule is not necessary. just use the quotient rule:

$\displaystyle \frac d{dx} \frac fg = \frac {f'g - fg'}{g^2}$

where $\displaystyle f,~g$ are functions of $\displaystyle x$

if you really want to employ the chain rule, you'd have to rewrite the expression as a product and use the product rule. you would need the chain rule to differentiate one of the factors. - Jan 10th 2010, 02:44 PMpickslides
You don't need the chain rule for this problem, you need the quoient rule.

- Jan 10th 2010, 02:54 PMtom@ballooncalculus
... which, in turn, is pretty much the chain rule...

http://www.ballooncalculus.org/asy/chain.png

... inside the product rule...

http://www.ballooncalculus.org/asy/prod.png

Straight continuous lines differentiate downwards (integrate up) with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).

So, just in case a picture helps...

http://www.ballooncalculus.org/asy/d...d/trigFrac.png

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