If 3 f(x)-2 fleft(frac{1}{x}right)=x,then f^{ | vedclass

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(B) Given equation is $3 f(x)-2 f\left(\frac{1}{x}\right)=x$ ...$(i)$ Replacing $x$ with $\frac{1}{x}$ in equation $(i)$,we get: $3 f\left(\frac{1}{x}

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$\frac{1}{2}$

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(B) Given equation is $3 f(x)-2 f\left(\frac{1}{x}\right)=x$ ...$(i)$ Replacing $x$ with $\frac{1}{x}$ in equation $(i)$,we get: $3 f\left(\frac{1}{x}\right)-2 f(x)=\frac{1}{x}$ ...(ii) To eliminate $f\left(\frac{1}{x}\right)$,multiply equation $(i)$ by $3$ and equation (ii) by $2$: $9 f(x)-6 f\left(\frac{1}{x}\right)=3 x$ ...(iii) $6 f\left(\frac{1}{x}\right)-4 f(x)=\frac{2}{x}$ ...(iv) Adding equation (iii) and (iv): $5 f(x) = 3 x + \frac{2}{x} \Rightarrow f(x) = \frac{3}{5} x + \frac{2}{5 x}$ Differentiating with respect to $x$: $f^{\prime}(x) = \frac{3}{5} - \frac{2}{5 x^2}$ Now,substitute $x=2$: $f^{\prime}(2) = \frac{3}{5} - \frac{2}{5(2^2)} = \frac{3}{5} - \frac{2}{20} = \frac{3}{5} - \frac{1}{10} = \frac{6-1}{10} = \frac{5}{10} = \frac{1}{2}$.

If $3 f(x)-2 f\left(\frac{1}{x}\right)=x$,then $f^{\prime}(2)=$

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