If 8 f(x)+6 fleft(frac{1}{x}right)=x+5 and y= | vedclass

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(C) Given equations are:$8 f(x)+6 f\left(\frac{1}{x}\right)=x+5$  --- $(i)$$y=x^2 f(x)$  --- $(ii)$Replace $x$ with $\frac{1}{x}$ in equation $(i)$:$8

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$\frac{-1}{14}$

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(C) Given equations are:$8 f(x)+6 f\left(\frac{1}{x}\right)=x+5$  --- $(i)$$y=x^2 f(x)$  --- $(ii)$Replace $x$ with $\frac{1}{x}$ in equation $(i)$:$8 f\left(\frac{1}{x}\right)+6 f(x)=\frac{1}{x}+5$  --- $(iii)$To eliminate $f\left(\frac{1}{x}\right)$,multiply $(i)$ by $4$ and $(iii)$ by $3$:$32 f(x)+24 f\left(\frac{1}{x}\right)=4x+20$$18 f(x)+24 f\left(\frac{1}{x}\right)=\frac{3}{x}+15$Subtracting the second from the first:$14 f(x)=4x-\frac{3}{x}+5$Multiply by $x^2$ to get $y$:$14 x^2 f(x)=4x^3-3x+5x^2$$14 y=4x^3+5x^2-3x$Differentiating with respect to $x$:$14 \frac{d y}{d x}=12x^2+10x-3$At $x=-1$:$14 \left(\frac{d y}{d x}\right)_{x=-1}=12(-1)^2+10(-1)-3 = 12-10-3 = -1$$\frac{d y}{d x} = -\frac{1}{14}$

If $8 f(x)+6 f\left(\frac{1}{x}\right)=x+5$ and $y=x^2 f(x)$,then $\frac{d y}{d x}$ at $x=-1$ equals

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