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

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To find the derivative of x^2+6x+10 by first principle.

Let y = f(x)= x^2+6x+10.

Then dy/dx = lt {f(+h)-f(x0}/h as h--> 0.

Therefore ,

dy/dx = lt {(x+h)^2+6(x+h)+10 - (x^2+6x+10)}/h as h--> 0.

dy/dx = lt {x^2+2hx+h^2+6x+6h+10-6x^2-6x-10}/h.

dy/dx = lt {2hx+h^2 +6h}/h.

dy/dx = lth{2x+6+h}/h.

dy/dx = lt {2x+6+h}.

dy/dx = 2x+6.

Therefore the derivative of x^2+6x+10 by first principle is 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**