If x neq 0 and f(x) satisfies 8 f(x) + 6 f(fr | vedclass

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(D) Given the equation: $8 f(x) + 6 f(\frac{1}{x}) = x + 5$ $(1)$ Replacing $x$ with $\frac{1}{x}$ in $(1)$,we get: $8 f(\frac{1}{x}) + 6 f(x) = \frac

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$\frac{19}{14}$

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(D) Given the equation: $8 f(x) + 6 f(\frac{1}{x}) = x + 5$ $(1)$ Replacing $x$ with $\frac{1}{x}$ in $(1)$,we get: $8 f(\frac{1}{x}) + 6 f(x) = \frac{1}{x} + 5$ $(2)$ Multiply $(1)$ by $4$ and $(2)$ by $3$: $32 f(x) + 24 f(\frac{1}{x}) = 4x + 20$ $18 f(x) + 24 f(\frac{1}{x}) = \frac{3}{x} + 15$ Subtracting the two equations: $(32 - 18) f(x) = 4x - \frac{3}{x} + 5$ $14 f(x) = 4x - \frac{3}{x} + 5 \Rightarrow f(x) = \frac{4x^2 + 5x - 3}{14x}$ Now,$x^2 f(x) = x^2 \left( \frac{4x^2 + 5x - 3}{14x} \right) = \frac{4x^3 + 5x^2 - 3x}{14}$ Differentiating with respect to $x$: $\frac{d}{dx} (x^2 f(x)) = \frac{1}{14} (12x^2 + 10x - 3)$ At $x = 1$: $\frac{d}{dx} (x^2 f(x)) = \frac{1}{14} (12(1)^2 + 10(1) - 3) = \frac{12 + 10 - 3}{14} = \frac{19}{14}$

If $x \neq 0$ and $f(x)$ satisfies $8 f(x) + 6 f(\frac{1}{x}) = x + 5$,then the value of $\frac{d}{dx} (x^2 f(x))$ at $x = 1$ is

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