Stuck on an FP2 question!

I understand how to do it for say y = (tan(1+x)^-1) but this fraction is really throwing me off... any advice?

Thanks, @zetamcfc, @the bear and @B_9710. I managed to get part a) with what you guys said but still struggling with part b). I even just had a look at the mark scheme and I still am not really understanding b)...


For the post above, where have you pulled the equation "y=arctanx + c" from? I understand it may give you the answer but I don't understand the logical steps?

What's part b?

Sorry, I didn't realise the photo didn't have the letters in there. Part b) is "Hence, given that x<1... etc."

Sorry, I didn't realise the photo didn't have the letters in there. Part b) is "Hence, given that x<1... etc."

Thanks, @zetamcfc, @the bear and @B_9710. I managed to get part a) with what you guys said but still struggling with part b). I even just had a look at the mark scheme and I still am not really understanding b)...


For the post above, where have you pulled the equation "y=arctanx + c" from? I understand it may give you the answer but I don't understand the logical steps?

We need to know what ...etc is

Ah, of course. Got it!! Thanks everybody

I put the question in the OP

I wouldn't do it the way that the other users have, I'd write $f(x) = \arctan \frac{1+x}{1-x} - \arctan x$, then from the first part, notice that:

$\displaystyle f'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \left(\arctan \frac{1+x}{1-x} - \arctan x \right) = \frac{1}{1+x^2} - \frac{1}{1+x^2} = 0$

So that $f'(x) = 0 \Rightarrow f(x) = c$. Then $f(0) = \arctan{1} - \arctan{0} = \frac{\pi}{4}$.

I wouldn't do it the way that the other users have, I'd write $f(x) = \arctan \frac{1+x}{1-x} - \arctan x$, then from the first part, notice that:

$\displaystyle f'(x) = \frac{\mathrm{d}}{\mathrm{d}x} \left(\arctan \frac{1+x}{1-x} - \arctan x \right) = \frac{1}{1+x^2} - \frac{1}{1+x^2} = 0$

So that $f'(x) = 0 \Rightarrow f(x) = c$. Then $f(0) = \arctan{1} - \arctan{0} = \frac{\pi}{4}$.

Why would you know that $f'(x)=0$?

I assume you're not used that fact $f(x)=\dfrac{\pi}{4}$ as that'd be circular reasoning.

STEP I question there?

Because $(\arctan x)' = \frac{1}{1+x^2}$ and $\left(\arctan \frac{1+x}{1-x}\right)' = \frac{1}{1+x^2}$.

Derivative is a linear operator so since you're differentiating the difference of two things whose derivatives are the same, then $f'(x) = 0$. So $f(x)$ is a constant and plugging in any value of $x$ will reveal the value of this constant.

That was stupid of me. Didn't notice they were the same. Thanks. Noted.

I think that was the intended solution as well - they gave you part (a) to show the derivative of that is the derivative of $\arctan x$ too. It's a nice concept.

http://filestore.aqa.org.uk/subjects/AQA-MFP2-W-MS-JUN14.PDF

Q7 if you want the desired solution