how do I differentiate arctan [(1-x)/(3+x)] ??

I know you can use the standard differential for arctan = 1/(1+x^2) so I subbed in (1-x)/(3+x) for x but now don't I have to multiply by the derivative of that? But the MS is giving me a different answer??

Please help x

Reply 1

old_engineer

Original post by olivier_

Your proposed method looks sound. I would suggest posting your working and explaining how the mark scheme differs from that.

Reply 2

olivier_

Original post by old_engineer

Your proposed method looks sound. I would suggest posting your working and explaining how the mark scheme differs from that.

Thank you for the reply. I hope you can see the attached image which has my working

Reply 3

olivier_

the derivative of the quotient doesn't simplify to what the mark scheme says

Reply 4

old_engineer

Original post by olivier_

the derivative of the quotient doesn't simplify to what the mark scheme says

The derivative you need here is the derivative of (1-x) / (3+x) ... i.e. the derivative of the inner function of the function of a function.

(edited 1 year ago)

Reply 5

olivier_

Original post by old_engineer

The derivative you need here is the derivative of (1-x) / (3+x) ... i.e. the derivative of the inner function of the function of a function.

ohh so just (1-x)/(3+x) and not [(1-x)/(3+x)]^2 because the square is part of the formula and you only need to multiply by the derivative of whatever you replace x with in the formula! I think I understand! Thank you xxx

Reply 6

old_engineer

Original post by olivier_

...so...you only need to multiply by the derivative of whatever you replace x with in the formula!

Yes. There are various ways of stating that in terms of mathematical notation, but if you consider the original function as f(x) = g(h(x)), then
f'(x) = (dg/dh)(dh/dx)

(edited 1 year ago)