# Math Find the derivative of x^2+6x+10 using the first principle.

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### 2 Answers

You need to find derivative of the given function from first principle, such that:

`(dy)/(dx) = lim_(Delta x -> 0) (f(x + Delta x) - f(x))/(Delta x)`

`(dy)/(dx) = lim_(Delta x -> 0) ((x+Delta x)^2+6(x + Delta x)+10 - x^2 - 6x - 10)/(Delta x)`

`(dy)/(dx) = lim_(Delta x -> 0) (x^2 + 2xDelta x + Delta^2 x + 6x + 6Delta x + 10 - x^2 - 6x - 10)/(Delta x)`

Reducing duplicate terms yields:

`(dy)/(dx) = lim_(Delta x -> 0) (2xDelta x + (Delta x)^2 + 6Delta x)/(Delta x)`

Factoring out `Delta x` yields:

`(dy)/(dx) = lim_(Delta x -> 0) Delta x*(2x + 6 + Delta x)/(Delta x)`

Reducing duplicate factors yields:

`(dy)/(dx) = lim_(Delta x -> 0) (2x + 6 + Delta x)`

`(dy)/(dx) = 2x + 6 + 0`

`(dy)/(dx) = 2x + 6`

**Hence, evaluating the derivatove of the given function, from the first principle, yields **`(dy)/(dx) = 2x + 6.`

First, we'll express the first principle of finding the derivative of a given function:

lim [f(x+h) - f(x)]/h, for h->0

We'll apply the principle to the given polynomial:

lim {[(x+h)^2 + 6(x+h) + 10]-(x^2 + 6x + 10)}/h

The next step is to expand the square:

lim {[(x^2+2xh + h^2) + 6x+6h + 10]-(x^2 + 6x + 10)}/h

We'll remove the brackets and combine and eliminate like terms:

lim (2xh + h^2 + 6h)/h

We'll factorize by h:

lim h(2x + h + 6)/h

We'll simplify and we'll get:

lim (2x + h + 6)

We'll substitute h by 0 and we'll get:

lim (2x + h + 6) = 2x + 6

So, the first derivative of the given function is:

**f'(x) = 2x + 6**