If 5 f(x) + 3 fleft(frac{1}{x}right) = x + 2 | vedclass

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(B) Given the equation: $5 f(x) + 3 f\left(\frac{1}{x}\right) = x + 2$  $(1)$Replace $x$ with $\frac{1}{x}$ in equation $(1)$: $5 f\left(\frac{1}{x}\r

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$\frac{7}{8}$

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(B) Given the equation: $5 f(x) + 3 f\left(\frac{1}{x}\right) = x + 2$  $(1)$Replace $x$ with $\frac{1}{x}$ in equation $(1)$: $5 f\left(\frac{1}{x}\right) + 3 f(x) = \frac{1}{x} + 2$  $(2)$Multiply $(1)$ by $5$ and $(2)$ by $3$: $25 f(x) + 15 f\left(\frac{1}{x}\right) = 5x + 10$$9 f(x) + 15 f\left(\frac{1}{x}\right) = \frac{3}{x} + 6$Subtract the second from the first: $(25 - 9) f(x) = 5x - \frac{3}{x} + 4$$16 f(x) = 5x - \frac{3}{x} + 4 \implies f(x) = \frac{5x^2 + 4x - 3}{16x}$Given $y = x f(x) = \frac{5x^2 + 4x - 3}{16}$Differentiating with respect to $x$: $\frac{dy}{dx} = \frac{1}{16} (10x + 4)$At $x = 1$: $\frac{dy}{dx} = \frac{1}{16} (10(1) + 4) = \frac{14}{16} = \frac{7}{8}$

If $5 f(x) + 3 f\left(\frac{1}{x}\right) = x + 2$ and $y = x f(x)$,then $\frac{dy}{dx}$ at $x = 1$ is equal to

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