Page:Calculus Made Easy.pdf/41

Doing the cubing we obtain

$y+dy=x^{3}+3x^{2}\cdot dx+3x(dx)^{2}+(dx)^{3}$.

Now we know that we may neglect small quantities of the second and third orders; since, when $dy$ and $dx$ are both made indefinitely small, $(dx)^{2}$ and $(dx)^{3}$ will become indefinitely smaller by comparison. So, regarding them as negligible, we have left:

$y+dy=x^{3}+3x^{2}\cdot dx$.

But $y=x^{3}$; and, subtracting this, we have:

$dy=3x^{2}\cdot dx$,

and

${\frac {dy}{dx}}=3x^{2}$.

Case 3.

Try differentiating $y=x^{4}$. Starting as before by letting both $y$ and $x$ grow a bit, we have:

$y+dy=(x+dx)^{4}$.

Working out the raising to the fourth power, we get

$y+dy=x^{4}+4x^{3}dx+6x^{2}(dx)^{2}+4x(dx)^{3}+(dx)^{4}$.

Then striking out the terms containing all the higher powers of $dx$, as being negligible by comparison, we have

$y+dy=x^{4}+4x^{3}dx$.

Subtracting the original $y=x^{4}$, we have left

$dy=4x^{3}dx$,

and

${\frac {dy}{dx}}=4x^{3}$.

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