$7f(x) + 5f\left( {\frac{1}{x}} \right) = x + 1$$t = 24xf(x)$${\left. {\frac{{dt}}{{dx}}} \right|_{x = - 1/7}} = ?$

We have, $7f(x) + 5f\left( {\frac{1}{x}} \right) = x + 1$ ..........(A)

Replacing x by $\frac {1}{x}$, $7f\left( {\frac{1}{x}} \right) + 5f(x) = \frac{1}{x} + 1$ ...........(B)

(A) + (B) yields, $12\left[ {f(x) + f\left( {\frac{1}{x}} \right)} \right] = x + \frac{1}{x} + 2$

$\therefore f(x) + f\left( {\frac{1}{x}} \right) = \frac{x}{{12}} + \frac{1}{{12x}} + \frac{1}{6}$ ..........(C)

(A) - (B) yields, $2\left[ {f(x) - f\left( {\frac{1}{x}} \right)} \right] = x - \frac{1}{x}$

$\therefore f(x) - f\left( {\frac{1}{x}} \right) = \frac{x}{2} - \frac{1}{{2x}}$ ..........(D)

(C) + (D) yields, $2f(x) = \frac{{7x}}{{12}} - \frac{5}{{12x}} + \frac{1}{6}$

$\therefore f(x) = \frac{{7x}}{{24}} - \frac{5}{{24x}} + \frac{1}{{12}}$

Now, $t = 24xf(x) = 7{x^2} - 5 + 2x$

$\therefore{\left. {\frac{{dt}}{{dx}}} \right|_{x =  - 1/7}} = {\left. {14x + 2} \right|_{x =  - 1/7}} = 0$