need help with the attachment

- anonymous

need help with the attachment

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- anonymous

##### 1 Attachment

- anonymous

a) Just plug in 1 for the numerator and g(x) for the denominator in the quotient rule.
b) \[ -{\frac {1+k{{\rm e}^{s}}}{ \left( s+k{{\rm e}^{s}} \right) ^{2}}} \] assuming y is a function of s.
c) Follows from \[ x^{-n} = \frac{1}{x^n} \] and \[ \frac{d}{dx} x^n = nx^{n-1} \]

- anonymous

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

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## More answers

- anonymous

What do you mean? The reciprocal rule is a special case of the quotient rule. You can use either one for part b, but it's easier to use the reciprocal rule.

- anonymous

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

- anonymous

but g(x) is just a variable

- anonymous

ok I got it

- anonymous

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

- anonymous

Ok, hit good answer.

- anonymous

OK I need help with differentiating the bottom on part could you show me step by step how to differentiate that one?

- anonymous

part b

- anonymous

and also the numerator of the function

- anonymous

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

- anonymous

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

- anonymous

What is the derivative of s?

- anonymous

##### 1 Attachment

- anonymous

1

- anonymous

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

- anonymous

@ta, see the attachee

- anonymous

e^(s)

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

1+e^(s)

- anonymous

No! Don't forget the factor k!

- anonymous

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

- anonymous

But you must not forget it :D

- anonymous

##### 1 Attachment

- anonymous

here u r

- anonymous

raheen there is still no explanation of part b :D

- anonymous

been answered all

- anonymous

part b asking to find using the rule only

- anonymous

ta123 do you have more questions?

- anonymous

whears the other 2 parts are to prove

- anonymous

@ta got that?

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

Your answer is correct ta.

- anonymous

yes, thanks for the help Y and raheen

- anonymous

ta why some ppl are using an offensive words?

- anonymous

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

- anonymous

yes

- anonymous

thank u.

- anonymous

no prob!

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