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Why are the derivatives of these functions the same? $$\frac{x^2+1}{x^2+x+1} \qquad\qquad \frac{-x}{x^2+x+1}$$

original exercise text (See part (e).)

I have tried to answer this question and consulted the answer booklet but this did not make much sense. Would someone be able to simplify or elaborate further? Thank you :)

The answer guide said the derivates are the same as differ at most by an additive constant.

Blue
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strugglingmathematician
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    Welcome to [math.se] SE. Take a [tour]. You'll find that simple "Here's the statement of my question, solve it for me" posts will be poorly received. What is better is for you to add context (with an [edit]): What you understand about the problem, what you've tried so far, etc.; something both to show you are part of the learning experience and to help us guide you to the appropriate help. You can consult this link for further guidance. – Another User Oct 22 '22 at 12:49
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    To check that the two expressions differ by a constant, subtract the second expression from the first and see what you get. –  Oct 22 '22 at 12:58
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    Consider what it means for two functions to have equal derivatives. Or vice versa, consider what distinguishes the possible antiderivatives of a function. – Jam Oct 22 '22 at 12:59
  • Remember that $\frac{d}{dx}$ is a linear operator as well – Doobius Oct 22 '22 at 13:00
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    Because if $f(x)-g(x)=Constant$ then........(You do know what). – Piquito Oct 22 '22 at 13:10
  • Add "1" to the expression on the right, and see what it looks like when you put it over the same denominator. – johnnyb Oct 22 '22 at 13:33

1 Answers1

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$$ {d\over dx} (f(x)+1)={df\over dx}+{d1\over dx}={df\over dx} $$ so we can add $1$ to a function and the derivative of the sum is the same as the derivative of the function. So $$ \begin{align} \frac{-x}{x^2+x+1}+1 &=\frac{-x}{x^2+x+1}+\frac{x^2+x+1}{x^2+x+1}\\ &={x^2+x+1-x\over x^2+x+1}\\ &={x^2+1\over x^2+x+1}\\ \end{align} $$

Suzu Hirose
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