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

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Aicharya · Accepted Answer

$8 f(x)+6 f\left(\frac{1}{x}ight)=x+5...(i)$ Differentiating w.r.t. $x$, we get $8 f^{\prime}(x)+6 f^{\prime}\left(\frac{1}{x}ight)\left(\frac{-1}{x^2}ight)=1...(ii)$ Substituting $x=-1$ in (i), we get $8 f(-1)+6 f(-1)=4$ $\begin{array}{ll} 	herefore & 14 f(-1)=4 \ 	herefore & f(-1)=\frac{4}{14}=\frac{2}{7}...(iii) \end{array}$ Substituting $x=-1$ in (ii), we get $\begin{aligned} & \mathrm{f}^{\prime}(-1)=\frac{1}{2} \. & y=x^2 \mathrm{f}(x) \end{aligned}$...(iv) $\begin{aligned} 	herefore \quad \frac{\mathrm{d} y}{\mathrm{~d} x} & =2 \ \left.	herefore \quad \frac{\mathrm{~d} y}{\mathrm{~d} x}ight|_{x=-1} & =2(-1)+x^2 \mathrm{f}^{\prime}(x) \ & =\frac{-4}{7}+\frac{1}{2} \quad \ldots[	ext { From (iii) and (iv) }] \ & =-\frac{1}{14}\end{aligned}$

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

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