Formulas for finding the differentials of functions

Since the differential of a function is its derivative multiplied by the differential of the independent variable, it follows at once that the formulas for finding differentials are the same as those for finding derivatives given in §5.1, if we multiply each one by $ dx$.

This gives us

I
$ d(c) = 0.$

II
$ d(x) = dx.$
III
$ d(u + v - w) = du + dv - dw.$
IV
$ d(cv) = cdv.$
V
$ d(uv) = udv + vdu$.
VI
$ d(v^n) = nv^{n - 1}dv$.
VIa
$ d(x^n) = nx^{n-1}dx$.
VII
$ d \left ( \frac{u}{v} \right ) = \frac{v du - u dv}{v^2}$.
VIIa
$ d \left ( \frac{u}{c} \right ) = \frac{du}{c}$.
VIII
$ d(\log\, av) = \log_a e \frac{dv}{v}$.
IX
$ d(a^v) = a^v\log\, a dv$.
IXa
$ d(e^v) = e^vdv$.
X
$ d(u^v) = vu^{v-1} du + \log u \cdot u^v \cdot dv$.
XI
$ d(\sin\, v) = \cos\, v dv$.
XII
$ d(\cos\, v) = -\sin\, vdv$.
XIII
$ d(\tan\,v) = \sec^2vdv$, etc.
XVIII
$ d(\arcsin\,v) = \frac{dv}{\sqrt{1 - v^2}}$, etc.

The term ``differentiation'' also includes the operation of finding differentials.

In finding differentials the easiest way is to find the derivative as usual, and then multiply the result by $ dx$.

Example 9.7.1   Find the differential of

$\displaystyle \ y = \frac{x + 3}{x^2 + 3}.
$

Solution. $ dy = d \left( \frac{x + 3}{x^2 + 3} \right)
= \frac{(x^2 + 3)d(x + 3) - (x + ...
... + 3)dx - (x + 3) 2x dx}{(x^2 + 3)^2}
= \frac{(3 - 6x - x^2)dx}{(x^2 + 3)^2}.
$

Example 9.7.2   Find $ dy$ from $ b^2x^2 - a^2y^2 = a^2b^2$.

Solution. $ 2b^2xdx - 2a^2ydy = 0$. Therefore, $ dy= \frac{b^2 x}{a^2 y} dx$.

Example 9.7.3   Find $ dy$ from $ \rho^2 = a^2 \cos 2\theta$.

Solution. $ 2\rho d\rho
= -a^2 \sin 2\theta \cdot 2d\theta$. Therefore, $ d\rho= -\frac{a^2 \sin 2\theta}{\rho} d\theta$.

Example 9.7.4   Find $ d[\arcsin(3t - 4t^3)]$.

Solution. $ d[ \arcsin (3t - 4t^3) ]
= \frac{d(3t - 4t^3)}{\sqrt{1 - (3t - 4t^3)^2}}
= \frac{3 dt}{\sqrt{1 - t^2}}$.

david joyner 2008-11-22