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

Differentiating the given expression, we get8f′(x)+6f′1x−1x2−1=0⇒ 8f′(1)+6f′(1)(−1)=1⇒ f′(1)=1/2Also dydx=2xf(x)+x2f′(x)so,  dydxx=1=2f(1)+f′(1)Putting x = 1 in the given equation, we obtain⇒ 8f(1)+6f(1)−1=514f(1)=6⇒ f(1)=37Hence      dydxx=1=2,37+12=1914