Question

If $f\left(x\right)=\sqrt{x^2+6x+9}$ then $f^{\prime }\left(x\right)$ is equal to-

• A. $1$ for $x<-3$
• B. $-1$ for $x<-3$
• C. $1$ for all $xϵR$
• D. None of these

Answer 1

Answer: (B)

$-1$ for $x<-3$

$f\left(x\right)=\sqrt{x^2+6x+9}$
$f^{\prime }\left(x\right)=\frac{2x+6}{2\sqrt{x^2+6x+9}}$
$=\frac{x+3}{\sqrt{x^2+6x+9}}=\frac{x+3}{\sqrt{{(x+3)}^2}}$
For $x<-3$
$f^{\prime }\left(x\right)=-1$