If x ≠ 0 and f x satisfies 8 f x + 6 f 1 x = x + 5 , then d d x x 2 f x at x = 1 is

We have,

$8f\left(x\right)+6f\left(\frac{1}{x}\right)=x+5\Rightarrow f\left(-1\right)=\frac{2}{7}$

Differentiating on both sides

$8f^{\text{'}}\left(x\right)-\frac{6}{x^2}{f}^{\text{'}}\left(\frac{1}{x}\right)=1$

Put $x=-1\Rightarrow f^{\text{'}}\left(-1\right)=\frac{1}{2}$

Now differentiating $y=x^{2}f\left(x\right)$ we get,

$\frac{dy}{dx}=2xf\left(x\right)+x^2{f}^{\text{'}}\left(x\right)$

at $x=-1,\frac{dy}{dx}=2\left(-1\right)\times \frac{2}{7}+{\left(-1\right)}^2\frac{1}{2}=-\frac{1}{14}$