Help with derivative involving trig

**billabong7329** · March 16th 2008, 06:02 PM

find the derivative of :

$(sin^2(x) + 1) / (cos^2(x) + 1)$

I tried just using a combination of quotient and chain rule, but I got some REALLY huge nasty answer. I'm thinking there is someway to simplify this, but not sure how.

Someone told me this was the answer, but I'm not sure how they got it :

$sin(2x) + (2sec(x)^2) * (tan(x))$

**topsquark** · March 16th 2008, 06:37 PM

Originally Posted by billabong7329

find the derivative of :

$(sin^2(x) + 1) / (cos^2(x) + 1)$

I tried just using a combination of quotient and chain rule, but I got some REALLY huge nasty answer. I'm thinking there is someway to simplify this, but not sure how.

Someone told me this was the answer, but I'm not sure how they got it :

$sin(2x) + (2sec(x)^2) * (tan(x))$

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

$= \frac{2sin(x)~cos^3(x) + 2sin(x)~cos(x) + 2sin^3(x)~cos(x) - 2sin(x)~cos(x)}{(cos^2(x) + 1)^2}$

$= \frac{2sin(x)~cos^3(x) + 2sin^3(x)~cos(x)}{(cos^2(x) + 1)^2}$

$= \frac{2(sin(x)~cos(x))(cos^2(x) + sin^2(x))}{(cos^2(x) + 1)^2}$

$= \frac{2sin(x)~cos(x)}{(cos^2(x) + 1)^2}$

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

Now, you could do a half-angle formula on this and get
$= \frac{sin(2x)}{4~cos^2 \left ( \frac{x}{2} \right )}$

And so on, but I can't see this getting into your friend's form.

-Dan

**mr fantastic** · March 16th 2008, 06:38 PM

Originally Posted by billabong7329

find the derivative of :

$(sin^2(x) + 1) / (cos^2(x) + 1)$

I tried just using a combination of quotient and chain rule, but I got some REALLY huge nasty answer. I'm thinking there is someway to simplify this, but not sure how.

Someone told me this was the answer, but I'm not sure how they got it :

$sin(2x) + (2sec(x)^2) * (tan(x))$

Can you show your working? Note that:

1. The derivative of $\sin^2 x$ is $2 \sin x \cos x = \sin(2x)$ .

2. The derivative of $\cos^2 x$ is $-2 \cos x \sin x = - \sin(2x)$ .

3. $\cos^2 x + \sin^2 x = 1 \Rightarrow \cos^2 x + \sin^2 x + 2 = 3$ .....

**billabong7329** · March 16th 2008, 06:59 PM

thanks topsquark, that is the answer I had gotten ( the first one you posted, not the simplification)

I submitted it to my online homework thing but still got it wrong. I think I just got it wrong because I didn't distribute the negative sign in the second half of the numerator.

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