Let f be a differentiable function such that 8f(x)+6f(1/x)−x=5(x≠0) and y=x2f(x) then dydx at x=1 is

Solution:

Differentiating the given expression, we get

$\begin{aligned} 8 f^{'} (x) + 6 f^{'} (\frac{1}{x}) (- \frac{1}{x^{2}}) - 1 = 0 \\ \Rightarrow 8 f^{'} (1) + 6 f^{'} (1) (- 1) = 1 \\ \Rightarrow f^{'} (1) = 1 / 2 \end{aligned}$

Also $\frac{d y}{d x} = 2 x f (x) + x^{2} f^{'} (x)$

so, ${\frac{d y}{d x}|}_{x = 1} = 2 f (1) + f^{'} (1)$

Putting $x = 1$ in the given equation, we obtain

$\begin{array}{l} \Rightarrow 8 f (1) + 6 f (1) - 1 = 5 \\ 14 f (1) = 6 \\ \Rightarrow f (1) = \frac{3}{7} \end{array}$

Hence ${\frac{d y}{d x}|}_{x = 1} = 2, \frac{3}{7} + \frac{1}{2} = \frac{19}{14}$

Let $f$ be $a$ differentiable function such that $8 f (x) + 6 f (1 / x) - x = 5 (x \neq 0)$ and $y = x^{2} f (x)$ then
$\frac{d y}{d x}$ at $x = 1$ is

Solution:

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