1. cosine derivative

Differentiate: y = $\frac {1 + cosx} {1- cosx}$

2. Originally Posted by Samantha
Differentiate: y = $\frac {1 + cosx} {1- cosx}$
Use the quotient rule:

$
f(x)=\frac{g(x)}{h(x)}
$

then:

$
f'(x)=\frac{g'(x)h(x)-g(x)h'(x)}{(h(x))^2}
$

here $g(x)=1 + \cos(x)$ and $h(x)=1 - \cos(x)$, and so:

$
g'(x)=-\sin(x)
$

and:

$
h'(x)=\sin(x)
$

Putting this all together:

$
f'(x)=\frac{-\sin(x)(1-\cos(x))-(1+\cos(x))\sin(x)}{(1-\cos(x))^2}
$

which you can simplify for yourself.

RonL

3. Originally Posted by Samantha
Differentiate: y = $\frac {1 + cosx} {1- cosx}$
$y = \frac {1 + cos(x)} {1- cos(x)}$

Using the quotient rule:
$y^{\prime} = \frac{-sin(x) \cdot (1 - cos(x)) - (1 + cos(x)) \cdot sin(x)}{(1 - cos(x))^2}$

$y^{\prime} = \frac{-sin(x) + sin(x) cos(x) - sin(x) - sin(x) cos(x)}{(1 - cos(x))^2}$

$y^{\prime} = \frac{-2sin(x)}{(1 - cos(x))^2}$

We can play around with the denominator, but I see no value in that. $(1 - cos(x))^2 = 1 - 2cos(x) + cos^2(x)$ which gives us no cancellations that I can see.

-Dan

4. Find dy/dx for y= $(x^3) {\sqrt x+1}$

5. Originally Posted by Samantha
Find dy/dx for y= $(x^3) \sqrt x+1$
Please use a new post for a new question.

$y= (x^3) \sqrt x+1$

The simple way:

$y= x^{7/2}+1$

So
$y^{\prime} = \frac{7}{2}x^{5/2}$

-Dan

6. Originally Posted by topsquark
Please use a new post for a new question.

$y= (x^3) \sqrt x+1$

The simple way:

$y= x^{7/2}+1$

So
$y^{\prime} = \frac{7}{2}x^{5/2}$

-Dan
I'm not sure if I wrote it right. It's suppose to be square root of x+1

7. $y=x^3\sqrt{x+1}$
$y'=(x^3)'\sqrt{x+1}+(x^3)\left(\sqrt{x+1}\right)'$
$3x^2\sqrt{x+1}+x^3 \cdot \frac{1}{2}\cdot \frac{1}{\sqrt{x+1}}$

8. I didn't get it =(

9. ThePerfectHacker used the multiplication rule:
$(fg)^{\prime} = f^{\prime} \cdot g + f \cdot g^{\prime}$

So we want to find the derivative of $x^3 \sqrt{x + 1}$
then $f(x) = x^3$ and $g(x) = \sqrt{x + 1}$.

The derivative of $\sqrt{x + 1}$ is $\frac{1}{2\sqrt{x + 1}}$.

So
$(x^3 \sqrt{x + 1})^{\prime} = (x^3)^{\prime} \cdot \sqrt{x + 1} + x^3 \cdot (\sqrt{x + 1})^{\prime}$

$(x^3 \sqrt{x + 1})^{\prime} = 3x^2 \sqrt{x + 1} + x^3 \cdot \frac{1}{2\sqrt{x + 1}}$

as TPH said.

-Dan

10. Hello, Samantha!

If you want to surprise/impress/terrify your teacher, use these identities:

. . $\sin^2\!\left(\frac{\theta}{2}\right) \: = \:\frac{1 - \cos\theta}{2} \quad\Rightarrow\quad 1 - \cos\theta \:=\:2\sin^2\!\left(\frac{\theta}{2}\right)$

. . $\cos^2\!\left(\frac{\theta}{2}\right) \: = \: \frac{1 + \cos\theta}{2}\quad\Rightarrow\quad1 + \cos\theta \:=\:2\cos^2\!\left(\frac{\theta}{2}\right)$

Differentiate: . $y \:= \:\frac {1 + \cos x} {1- \cos x}$

We have: . $y\;=\;\frac{1 + \cos x}{1 - \cos x} \;=\;\frac{2\cos^2\!\left(\frac{x}{2}\right)}{2\si n^2\!\left(\frac{x}{2}\right)} \;= \;\cot^2\!\left(\frac{x}{2}\right)$

Then: . $y' \;=\;2\cot\left(\frac{x}{2}\right)\cdot\left[-\csc^2\!\left(\frac{x}{2}\right)\right] \cdot\frac{1}{2}$

. . Therefore: . $\boxed{\;y' \;= \;-\csc^2\!\left(\frac{x}{2}\right)\cot\left(\frac{x} {2}\right)}$

11. Originally Posted by Soroban
Hello, Samantha!

If you want to surprise/impress/terrify your teacher, use these identities:

. . $\sin^2\!\left(\frac{\theta}{2}\right) \: = \:\frac{1 - \cos\theta}{2} \quad\Rightarrow\quad 1 - \cos\theta \:=\:2\sin^2\!\left(\frac{\theta}{2}\right)$

. . $\cos^2\!\left(\frac{\theta}{2}\right) \: = \: \frac{1 + \cos\theta}{2}\quad\Rightarrow\quad1 + \cos\theta \:=\:2\cos^2\!\left(\frac{\theta}{2}\right)$

We have: . $y\;=\;\frac{1 + \cos x}{1 - \cos x} \;=\;\frac{2\cos^2\!\left(\frac{x}{2}\right)}{2\si n^2\!\left(\frac{x}{2}\right)} \;= \;\cot^2\!\left(\frac{x}{2}\right)$

Then: . $y' \;=\;2\cot\left(\frac{x}{2}\right)\cdot\left[-\csc^2\!\left(\frac{x}{2}\right)\right] \cdot\frac{1}{2}$

. . Therefore: . $\boxed{\;y' \;= \;-\csc^2\!\left(\frac{x}{2}\right)\cot\left(\frac{x} {2}\right)}$

(Chuckles) I like it.

-Dan