Thread: I missed a class on one of the Derivative Rules.

1. I missed a class on one of the Derivative Rules.

Just explain how one would go about doing this:

What's the rule to this?

2. Originally Posted by Zanderist
Just explain how one would go about doing this:

What's the rule to this?
You need to use the Quotient Rule:

If $f(x) = \frac{g(x)}{h(x)}$ then $f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}$.

3. to solve that you would use th quotient rules as well as the sin and cos rules.

$
\frac{dy}{dx}\sin{x}=\cos{x}
$

$
\frac{dy}{dx}\cos{x}=-\sin{x}
$

so using the quotient rule, $\frac{dy}{dx}\frac{u}{v}=\frac{vu'-uv'}{v^2}$

$
u=4\sin{x}
$

$
v=2+\cos{x}
$

$
u'=4\cos{x}
$

$
v'=-\sin{x}
$

fill it all in and solve

4. Originally Posted by Zanderist
Just explain how one would go about doing this:

What's the rule to this?
this would be:

4cosx(2+cosx) - 4sinx(-sinx)
----------------------------
(2+cosx)^2

You can simplify even if you miss the entire semester. Go for it!!

5. Originally Posted by harish21
this would be:

4cosx(2+cosx) - 4sinx(-sinx)
----------------------------
(2+cosx)^2

You can simplify even if you miss the entire semester. Go for it!!
Yes, now simplify it.

6. tan(x)-5
--------
sec(x)

This some how turns into,

-(tan(x)^2-5*tan(x)-sec(x)^2)
----------------------------------
sec(x)

I don't see how the quotient rule gives me that.

7. Originally Posted by Zanderist
tan(x)-5
--------
sec(x)

This some how turns into,

-(tan(x)^2-5*tan(x)-sec(x)^2)
----------------------------------
sec(x)

I don't see how the quotient rule gives me that.
I would actually have done:

$\frac{8\cos{x} + 4\cos^2{x} + 4\sin^2{x}}{(2 + \cos{x})^2}$

$= \frac{8\cos{x} + 4(\cos^2{x} + \sin^2{x})}{(2 + \cos{x})^2}$

$= \frac{8\cos{x} + 4}{(2 + \cos{x})^2}$

$= \frac{4(2\cos{x} + 1)}{(2 + \cos{x})^2}$.

8. Originally Posted by Zanderist
tan(x)-5
--------
sec(x)

This some how turns into,

-(tan(x)^2-5*tan(x)-sec(x)^2)
----------------------------------
sec(x)

I don't see how the quotient rule gives me that.
tan(x)-5 / secx = (sinx - 5cosx) [ Reason: tanx=sinx/cosx and secx=1/cosx]

Derivate this and you get cosx + 5 sinx