find $f'(x)$ for $f(x)= \frac{x+5}{x+3}$ and $f(x) = \frac{1}{x^2 + 1}$

Question

Find $f'(x)$ for $f(x)=\frac{x+5}{x+3}$ and $f(x)=\frac{1}{x^2 +1}$ and determine all values $c$ where $f'(c)=0$.

I used the quotient rule to find the derivatives. Then I let it to be equal to $0$.

For the first one, I get $$f'(x)=-\frac{2}{(x+3)^2}$$ and when I let it equal to $0$, I get -2/0 for x=-3. but this is not defined

I am stuck with second equation as I get $$f'(x)=-\frac{2x}{(x^2+1)^2}$$

Answer 1

For the first we have $$\left(1+\frac{2}{x+3}\right)'=-\frac{2}{(x+3)^2}\neq0$$ and for the second, which was $1+\frac{1}{x^2}$it's also impossible.

If you mean the the second is $\frac{1}{1+x^2}$ then we get $-\frac{2x}{(x^2+1)^2},$ which gives $c=0$.

Answer 2

for $$f(x)=\frac{x+5}{x+4}$$ we obtain by the Quotient rule $$f'(x)=\frac{x+3-(x+5)}{(x+3)^2}$$ simplifying we obtain $$f'(x)=-\frac{2}{(x+3)^2}$$ this can not be Zero for any $x$ for the second we get $$f'(x)=\frac{-2x}{(x^2+1)^2}$$ and $f'(x)=0$ if $x=0$