Derivatives: why does the integer disappear?

I'm new to this so don't slaughter me.

x^n -> nx^(n-1)

So you do that throughout, but in an example I was looking at,

x^2 +5x -6

becomes

2x + 5

I get the 5 and 2x part, but how can the 6 go, the power becomes zero, but that will turn the 1x-6 into a 1 ? Clearly not, what am I missing?

Reply 1

Math12345

14

6=6*1=6x^0

Differentiate:

0*6x^{-1} = 0

(edited 7 years ago)

Reply 2

SuchBants

OP

8

Original post by Math12345

6=6*1=6x^0

Differentiate:

0*6x^{-1} = 0

Thank you

Reply 3

EnglishMuon

16

Original post by SuchBants

I'm new to this so don't slaughter me.

x^n -> nx^(n-1)

So you do that throughout, but in an example I was looking at,

x^2 +5x -6

becomes

2x + 5

I get the 5 and 2x part, but how can the 6 go, the power becomes zero, but that will turn the 1x-6 into a 1 ? Clearly not, what am I missing?

A good way to understand it is to look at the geometrical interpretation:
Consider a straight line graph

y=mx+c

, then

\dfrac{dy}{dx}

is the gradient of that graph (or any graph) at a given point x. If we vary c, i.e. change it to be larger or smaller, we just translate the graph up parallel to the y axis hence we are not effecting the gradient i.e. the +c cant affect the derivative.

Reply 4

Zacken

22

In accordance with the above - the derivative measures the gradient of a curve at any given point, and the gradient of the line y = c is 0 (it's just a straight line).

Derivatives: why does the integer disappear?

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