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

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(A) Given equation is $3 f(x)-2 f\left(\frac{1}{x}\right)=x$ ...$(i)$Replace $x$ with $\frac{1}{x}$ in Eq. $(i)$:$3 f\left(\frac{1}{x}\right)-2 f(x)=\

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$\frac{1}{2}$

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

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

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