Page:Calculus Made Easy.pdf/42

Now all these cases are quite easy. Let us collect the results to see if we can infer any general rule. Put them in two columns, the values of $y$ in one and the corresponding values found for ${\frac {dy}{dx}}$ in the other: thus

$y$	${\frac {dy}{dx}}$
$x^{2}$	$2x$
$x^{3}$	$3x^{2}$
$x^{4}$	$4x^{3}$

Just look at these results: the operation of differentiating appears to have had the effect of diminishing the power of $x$ by $1$ (for example in the last case reducing $x^{4}$ to $x^{3}$), and at the same time multiplying by a number (the same number in fact which originally appeared as the power). Now, when you have once seen this, you might easily conjecture how the others will run. You would expect that differentiating $x^{5}$ would give $5x^{4}$, or differentiating $x^{6}$ would give $6x^{5}$. If you hesitate, try one of these, and see whether the conjecture comes right.

Try $y=x^{5}$.

Then

$y+dy=$	$(x+dx)^{5}$
$=$	$x^{5}+5x^{4}dx+10x^{3}(dx)^{2}+10x^{2}(dx)^{3}+5x(dx)^{4}+(dx)^{5}$.

Neglecting all the terms containing small quantities of the higher orders, we have left

$y+dy=x^{5}+5x^{4}dx$,