Question

If $y=f\left(x\right)$ is an odd differentiable function defined on $\left(-\infty ,\infty \right)$ such that $f^{{}^{\prime }}\left(3\right)=-2$ , then $\left|f^{{}^{\prime }}\left(-3\right)\right|$ equals to

Answer 1

Answer: 2

Given that $f\left(x\right)$ is an odd differentiable function which is defined for $x\in R$ .
From the property of odd functions, we can write,
$f\left(x\right)+f\left(-x\right)=0$
$f\left(x\right)=-f\left(-x\right)...\left(i\right)$
Differentiating equation $\left(i\right)$ with respect to $x$ , we get:
$f^{{}^{\prime }}\left(x\right)=-f^{{}^{\prime }}\left(-x\right)\frac{d}{dx}\left(-x\right)$
$f^{{}^{\prime }}\left(x\right)=-f^{{}^{\prime }}\left(-x\right)\left(-1\right)$
$f^{{}^{\prime }}\left(x\right)=f^{{}^{\prime }}\left(-x\right)$
Putting $x=3$ in above equation, we get:
$f^{{}^{\prime }}\left(3\right)=f^{{}^{\prime }}\left(-3\right)=-2$