Concept:

The rate of change of the value of a function f(x)with respect to a variable t, is given by:\(\rm \frac {df(x)}{dt}\)

Calculation:

To Find:Rate of change of \(\rm \sqrt {{x^2} + 48} \) with respect to x2

Let u be\(\rm \sqrt {{x^2} + 48} \)and v be x2

\(\rm \frac{{du}}{{dx}} = \frac{{1 \times 2x}}{{2\sqrt {{({x^2}} + 48)}}} = \frac{x}{{\sqrt {{({x^2}} + 48)}}}\)

\(\rm \frac{{dv}}{{dx}} = 2x\)

\(\rm \frac{{du}}{{dv}} = \;\frac{{\frac{{du}}{{dx}}}}{{\frac{{dv}}{{dx}}}} = \;\frac{{\frac{x}{{\sqrt {{({x^2}} + 48)}}}}}{{2x}} = \;\frac{x}{{\sqrt {{({x^2}} + 48)}}} \times \frac{1}{{2x}}\;\)

\(\rm \frac{{du}}{{dv}} = \frac{1}{{2\sqrt {{({x^2}} + 48)}}}\)

At x = 4

\(\rm \frac{{du}}{{dv}} = \frac{1}{{2\sqrt {{({4^2}} + 48)}}}\)

\(\rm \frac{{du}}{{dv}} = \frac{1}{{2\sqrt {64}}} = \frac{1}{{16}}\)

∴\(\rm \frac{{du}}{{dv}} = \frac{1}{{16}}\)