If $$4\sqrt 5f(x) + \sqrt3f\left (\frac {1}{x}\right) =\sqrt3 x+\sqrt5\\\text { and } y=xf (x)\text { for all values of x}$$
Find $\frac{dy}{dx} $ and show that $\left (\frac {dy}{dx}\right)_{x=2}=\frac {1}{77} [15\sqrt{15} +20]$
$\pmb {\text {The problem i face ;}}$
Everytime i substitute for $f (x)$ & $f (1/x)$ and differentiate with respect to $x$, there is always a $y$ in the answer.Can anyone help me out.I cant seem to get a specific value for $y$ using the given relationships neither can i get $y$ out of the answer.
Given that $$4\sqrt 5f(x) + \sqrt3f\left (\frac {1}{x}\right) =\sqrt3 x+\sqrt5$$ and replacing x by 1/x in the above expression we get $$4\sqrt 5f(1/x) + \sqrt3f(x) =\sqrt3/ x+\sqrt5.$$ Now multiply the first expression by $4\sqrt 5$ and second expression by $\sqrt3$ and subtract one from the other to eliminate the term $f(1/x).$
You get the explicit expression for $xf(x)=\dfrac{4\sqrt {15}x^2-3+20x-\sqrt {15}x}{77}$.
Now differentiate $xf(x)$ w.r.t.x and get the value at x=2. I think you can complete the answer!