To differentiate the function [tex]\( f(x) = 2x^3 + 3 \)[/tex] from first principles, we will use the definition of the derivative. The derivative of the function [tex]\( f(x) \)[/tex] at a point [tex]\( x \)[/tex] is given by the limit:
[tex]\[
f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}
\][/tex]
Let's go through the steps to find this derivative:
1. Express [tex]\( f(x + h) \)[/tex]:
Substitute [tex]\( x + h \)[/tex] into [tex]\( f(x) \)[/tex]:
[tex]\[
f(x + h) = 2(x + h)^3 + 3
\][/tex]
2. Expand [tex]\( (x + h)^3 \)[/tex]:
Use the binomial theorem to expand [tex]\( (x + h)^3 \)[/tex]:
[tex]\[
(x + h)^3 = x^3 + 3x^2h + 3xh^2 + h^3
\][/tex]
So,
[tex]\[
f(x + h) = 2(x^3 + 3x^2h + 3xh^2 + h^3) + 3 = 2x^3 + 6x^2h + 6xh^2 + 2h^3 + 3
\][/tex]
3. Compute [tex]\( f(x + h) - f(x) \)[/tex]:
Subtract [tex]\( f(x) \)[/tex] from [tex]\( f(x + h) \)[/tex]:
[tex]\[
f(x + h) - f(x) = (2x^3 + 6x^2h + 6xh^2 + 2h^3 + 3) - (2x^3 + 3)
\][/tex]
Simplify this expression:
[tex]\[
f(x + h) - f(x) = 6x^2h + 6xh^2 + 2h^3
\][/tex]
4. Form the difference quotient:
Divide [tex]\( f(x + h) - f(x) \)[/tex] by [tex]\( h \)[/tex]:
[tex]\[
\frac{f(x + h) - f(x)}{h} = \frac{6x^2h + 6xh^2 + 2h^3}{h}
\][/tex]
Simplify by canceling [tex]\( h \)[/tex]:
[tex]\[
\frac{f(x + h) - f(x)}{h} = 6x^2 + 6xh + 2h^2
\][/tex]
5. Take the limit as [tex]\( h \)[/tex] approaches 0:
Evaluate the limit of the difference quotient as [tex]\( h \)[/tex] approaches 0:
[tex]\[
f'(x) = \lim_{h \to 0} (6x^2 + 6xh + 2h^2)
\][/tex]
Since [tex]\( 6xh \)[/tex] and [tex]\( 2h^2 \)[/tex] both approach 0 as [tex]\( h \)[/tex] approaches 0, the limit simplifies to:
[tex]\[
f'(x) = 6x^2
\][/tex]
Therefore, the derivative of [tex]\( f(x) = 2x^3 + 3 \)[/tex] using first principles is:
[tex]\[
f'(x) = 6x^2
\][/tex]
