Page:Calculus Made Easy.pdf/174

Differentiate

(6) ${\displaystyle y=\log _{\epsilon }x^{n}}$.

(7) ${\displaystyle y=3\epsilon ^{-{\frac {x}{x-1}}}}$.

(8) ${\displaystyle y=(3x^{2}+1)\epsilon ^{-5x}}$.

(9) ${\displaystyle y=\log _{\epsilon }(x^{a}+a)}$.

(10) ${\displaystyle y=(3x^{2}-1)({\sqrt {x}}+1)}$.

(11) ${\displaystyle y={\dfrac {\log _{\epsilon }(x+3)}{x+3}}}$.

(12) ${\displaystyle y=a^{x}\times x^{a}}$.

(13) It was shown by Lord Kelvin that the speed of signalling through a submarine cable depends on the value of the ratio of the external diameter of the core to the diameter of the enclosed copper wire. If this ratio is called ${\displaystyle y}$, then the number of signals ${\displaystyle s}$ that can be sent per minute can be expressed by the formula

where ${\displaystyle a}$ is a constant depending on the length and the quality of the materials. Show that if these are given, ${\displaystyle s}$ will be a maximum if ${\displaystyle y=1\div {\sqrt {\epsilon }}}$.

(15) Differentiate ${\displaystyle y=\log _{\epsilon }(ax\epsilon ^{x})}$.

(16) Differentiate ${\displaystyle y=(\log _{\epsilon }ax)^{3}}$.

Let us return to the curve which has its successive ordinates in geometrical progression, such as that represented by the equation ${\displaystyle y=bp^{x}}$.

We can see, by putting ${\displaystyle x=0}$, that ${\displaystyle b}$ is the initial height of ${\displaystyle y}$.

Then when