RaySix

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at October 31st 2023, 5:08:18 pm.

Let $f(x) = x^3 - 2x^2 + 5x - 3$ be a differentiable function. Find the derivative of $f(x)$ using the definition of the derivative.

The derivative of a function $f(x)$ at a point $x = a$ is defined as:

f'(a) = \lim_{{h \to 0}} \frac{{f(a+h) - f(a)}}{h}

To find the derivative of $f(x) = x^3 - 2x^2 + 5x - 3$ , we need to evaluate this limit. Let's start by applying the definition of the derivative.

f'(a) = \lim_{{h \to 0}} \frac{{f(a+h) - f(a)}}{h} = \lim_{{h \to 0}} \frac{{(a+h)^3 - 2(a+h)^2 + 5(a+h) - 3 - (a^3 - 2a^2 + 5a - 3)}}{h}

Expanding the numerator, we have:

f'(a) = \lim_{{h \to 0}} \frac{{a^3 + 3a^2h + 3ah^2 + h^3 - 2(a^2 + 2ah + h^2) + 5(a+h) - 3 - a^3 + 2a^2 - 5a + 3}}{h}

Notice that many terms will cancel out:

f'(a) = \lim_{{h \to 0}} \frac{{3a^2h + 3ah^2 + h^3 - 2ah - 2h^2 + 5h}}{h}

Simplifying further:

f'(a) = \lim_{{h \to 0}} \frac{{h(3a^2 + 3ah + h^2 - 2a - 2h + 5)}}{h}

As $h$ approaches 0, we can cancel out the $h$ from the numerator and denominator:

f'(a) = \lim_{{h \to 0}} (3a^2 + 3ah + h^2 - 2a - 2h + 5)

Since all terms in the limit are continuous, we can substitute $h = 0$ directly:

f'(a) = 3a^2 + 0 + 0 - 2a - 0 + 5

Simplifying further, we get:

f'(a) = 3a^2 - 2a + 5

Therefore, the derivative of the function $f(x) = x^3 - 2x^2 + 5x - 3$ is $f'(x) = 3x^2 - 2x + 5$ .