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is the quotient rule the same thing as the reciprocal part b?

The quotient rule f'g-fg'/g^(2)

but g(x) is just a variable

ok I got it

What is your question? g is not just a variable, it's a function.

Ok, hit good answer.

part b

and also the numerator of the function

You need to differentiate\[ \Large s+k\cdot e^s \] with respect to s.

The derivative of a sum is the sum of derivatives, so differentiate s and then differentiate k*e^s.

What is the derivative of s?

Yes. And what is the derivative of e^s ?

e^(s)

Your attachment doesn't include part b) explanation, thats what he asked about.

Yes, so now you can add them together and you get the derivative

And then just plug that into the formula in part a)

1+e^(s)

No! Don't forget the factor k!

k does not depend on s, so you can just leave it unchanged

But you must not forget it :D

here u r

raheen there is still no explanation of part b :D

been answered all

part b asking to find using the rule only

ta123 do you have more questions?

whears the other 2 parts are to prove

lol raheen are you just trying to get a medal, because I can give you one if thats what you want :D

your answer is wrong raheen, I got -1-ke^(s)/(s+ke^(s))^2

no YThe, I always try to help , pls don read the things in wrong way

Your answer is correct ta.

yes, thanks for the help Y and raheen

ta why some ppl are using an offensive words?

I think I stayed with u b4 more than 2 hrs, without any annyance, right?

yes

thank u.

no prob!