1. ## derivatives

how do you get the derivative of:
$(\frac {1-\sin}{1+\cos})^2$
*hopefully i used the LaTex Does Correctly...

i tried following my teacher in this problem but easily got lost...
y = $2\sqrt{x-1} \csc^- \sqrt{x}$

and how would i find the normal and tangent line of:
$2x^2y + 2y^2=x^3 + 1$ at $(-1,1)$?
i know i have to take the derivative of the equation, but then what?

2. Originally Posted by >_<SHY_GUY>_<
how do you get the derivative of:
$(\frac {1-\sin}{1+\cos})^2$
*hopefully i used the LaTex Does Correctly...
Hi

Let $u(x) = \frac {1-\sin x}{1+\cos x}$

The derivative of $u^2$ is $2\:u(x)\:u'(x)$

The quotient rule is used to calculate u'(x)

$u'(x) = \frac{-\cos x\1+\cos x) - (- \sin x)\1-\sin x)}{(1+\cos x)^2} = \frac{-1 -\cos x + \sin x}{(1+\cos x)^2}" alt="u'(x) = \frac{-\cos x\1+\cos x) - (- \sin x)\1-\sin x)}{(1+\cos x)^2} = \frac{-1 -\cos x + \sin x}{(1+\cos x)^2}" />

The derivative of $\left(\frac {1-\sin x}{1+\cos x}\right)^2$ is therefore

$2\:\frac {1-\sin x}{1+\cos x}\:\:\frac{-1 -\cos x + \sin x}{(1+\cos x)^2}$

3. Great Work...i messed up in accidentally ignoring the negative sign when taking the derivative of $-\sin x$
thanks!

Originally Posted by running-gag
Hi

Let $u(x) = \frac {1-\sin x}{1+\cos x}$

The derivative of $u^2$ is $2\:u(x)\:u'(x)$

The quotient rule is used to calculate u'(x)

$u'(x) = \frac{-\cos x\1+\cos x) - (- \sin x)\1-\sin x)}{(1+\cos x)^2} = \frac{-1 -\cos x + \sin x}{(1+\cos x)^2}" alt="u'(x) = \frac{-\cos x\1+\cos x) - (- \sin x)\1-\sin x)}{(1+\cos x)^2} = \frac{-1 -\cos x + \sin x}{(1+\cos x)^2}" />

The derivative of $\left(\frac {1-\sin x}{1+\cos x}\right)^2$ is therefore

$2\:\frac {1-\sin x}{1+\cos x}\:\:\frac{-1 -\cos x + \sin x}{(1+\cos x)^2}$

4. To your second question, the derivative of your original function gives you the slope of the line tangent to that point.

Your tangent line will be in the form y = ax + b. Where a is the slope you got from your derivative at (-1, 1)

Say that slope was 2. Now, to calculate b, use the point you were given.

1 = 2(-1) + b

b = 3

So you're tangent line would be y = 2x + 3.

And the normal line is just the inverse of the tangent.

5. then i wouldnt have to worry about $2y^2$being $4y$y prime?

Originally Posted by derfleurer
To your second question, the derivative of your original function gives you the slope of the line tangent to that point.

Your tangent line will be in the form y = ax + b. Where a is the slope you got from your derivative at (-1, 1)

Say that slope was 2. Now, to calculate b, use the point you were given.

1 = 2(-1) + b

b = 3

So you're tangent line would be y = 2x + 3.

And the normal line is just the inverse of the tangent.

6. Nope. You only use the derivative to find the slope at that point. After that, you're only interested in the original function.

$2x^2y + 2y^2=x^3$

For the slope, derive implicitly

$4xy + 2x^2y' + 4yy' = 3x^2$

$y'(2x^2 + 4y) = 3x^2 - 4xy$

$y' = \frac{3x^2 - 4xy}{2x^2 + 4y}$

7. ok, now i see my mistake...i didnt know that the product rule applied to $2x^2y$i see why i get some problems wrong...thank you!

Originally Posted by derfleurer
Nope. You only use the derivative to find the slope at that point. After that, you're only interested in the original function.

$2x^2y + 2y^2=x^3$

For the slope, derive implicitly

$4xy + 2x^2y' + 4yy' = 3x^2$

$y'(2x^2 + 4y) = 3x^2 - 4xy$

$y' = \frac{3x^2 - 4xy}{2x^2 + 4y}$

8. A lot of people slip up on implicit differentiation. If you want a tip

Rewrite the question so that everything is on one side of the equals sign.

$2x^2y + 2y^2=x^3$

becomes

$2x^2y + 2y^2 - x^3=0$

Now consider this its own function.

$z = 2x^2y + 2y^2 - x^3$

Take the derivative of z with respect to x. This means considering all the y as just constants. So just like the derivative of $4x^2$ yeilds $8x$, the derivative of $4x^2y$ yields $8xy$:

$\frac{dz}{dx} = 4xy - 3x^2$

Now do the same for y

$\frac{dz}{dy} = 2x^2 + 4y$

Now your derivative will be in the form $\frac{-(dz/dx)}{dz/dy}$

$\frac{-(4xy - 3x^2)}{2x^2 + 4y}$

9. id say thats a really good trick there...thank you for that...i really appreciate it

ok, so i substitute (-1,-1) in the derivative of the equation to get the slope, then in the original equation for the b-intercept or how does that work? i get confused in finding the b-intercept...