# Math Help - Change in x and differentials

1. ## Change in x and differentials

2. f ' (x)= ( f(x)-f(xo) )/ (x-xo)

this implies..

f(x)-f(xo) = f ' (x) (x-xo)

3. Hello, zaurm!

Use differentials to approximate the change in
$f(x) \:=\:4x^5 - 6x^4 + 3x^2 - 5$ .if $x$ changes from 1 to 1.02
We have: . $x = 1,\:dx \:=\:0.02$

. . And: . $df \;=\;(20x^4 - 24x^3 + 6x)\,dx$

Therefore: . $df \;=\;\bigg[20(1^4) - 24(1^3) + 6(1) - 5\bigg]\,(0.02) \;=\;\boxed{0.04}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

I like to check my answer by calculating the actual change.

. . $f(1) \;=\;4(1^5) - 6(1^4) + 3(1^2) - 5 \;=\;-4$

. . $f(1.02) \;=\;4(1.02^5) - 6(1.02^4) + 3(1.02^2) - 5 \;=\;-3.957069747$

The actual change is: . $\Delta f \;=\;-3.957069747 - (-4) \;=\;{\color{blue}0.04}2930253$

. . . . Close enough!