# Use the first principle to determine the derivative of the function. f(x) = 5x^3 + 3x^2 - 2x + 15

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According to the first principle of derivation, the derivative of f(x) or f'(x)= lim(h->0){[f(x+h)-f(x)]/h}

Here f(x)=5x^3+3x^2-2x+15

f(x+h)=5(x+h)^3+3(x+h)^2-2(x+h)+15

=5*x^3+5*h^3+15*x^2*h+15*x*h^2+3*x^2+3*h^2+6*x*h-2x-2h+15

f(x+h)-f(x)=5*x^3+5h^3+15x^2*h+15x*h^2+3x^2+3h^2+6xh-2x-2h+15-5x^3+3x^2-2x+15

=5h^3+15x^2h+15xh^2+3h^2+6xh-2h

[f(x+h)-f(x)]/h=(5h^3+15x^2*h+15x*h^2+3h^2+6xh-2h)/h

=5h^2+15x^2+15xh+3h+6x-2

lim(h->0)[5h^2+15x^2+15xh+3h+6x-2]

=15x^2+6x-2

**Therefore (5x^3+3x^2-2x+15)' by the first principle is 15x^2+6x-2.**

We'll consider the function above as a sum of the following elementary functions:

- (5x^3) - is a power function and it's derivative will be calculated using the power rule. We know that (x^n) is a power function type, and it's derivative is (x^n)'=n*[x^(n-1)].We'll do the same way with (5x^3), where (5x^3)'=5*3*x^(3-1)=15x^2;
- (3x^2) - is also a power function and we'll calculate it's derivative as above:

(3x^2)'=3*2*x^(2-1)=6x;

- (-2x) - is a linear function, where it's first derivative is (2-x)'=-2. (-2x) can be as well considered as a power function where the power of the unknown x, is equal to 1. In each power function , the rule of finding out the derivative is: (x^n)=x*x*x*...*x. As we can se. it's all about a product of n factors. The rule in this case is: (x^n)'=x'*x^(n-1)+x*x'*x*...+....x*x*...*x'=n*x^(n-1) So,(-x)'=(-1)'*x+(-1)*(x)'=0*x+(-1)*1=-1
- (+15)- is a constant function and it's derivative is 0.

f'(x) = (5x^3 + 3x^2 - 2x + 15)'

f'(x) = (5x^3)'+(3x^2)'+( - 2x)' + (15)'

**f'(x) = **15x^2 + 6x - 2