Post

Created by @nathanedwards
 at October 31st 2023, 9:11:49 pm.

Exam Question:

Let f(x) be a differentiable function. Define the derivative of the function f(x) at a point x = a as:

f(a)=limh0f(a+h)f(a)h f'(a) = \lim_{{h \to 0}} \frac{{f(a + h) - f(a)}}{h}

a) Explain in plain English what the derivative of a function represents.

b) Find the derivative f'(x) of the function f(x) = 3x^2 + 2x - 1.

c) Evaluate the derivative f'(2) using the definition of the derivative.

Answer:

a) The derivative of a function represents the rate at which the function is changing at a specific point. In other words, it gives us the slope of the function at that point. It can indicate if the function is increasing or decreasing, and the steepness of the curve.

b) To find the derivative f'(x) of the function f(x) = 3x^2 + 2x - 1, we can apply differentiation rules.

Using the power rule, the derivative of x^n is nx^(n-1):

f(x)=ddx(3x2)+ddx(2x)ddx(1) f'(x) = \frac{{d}}{{dx}}(3x^2) + \frac{{d}}{{dx}}(2x) - \frac{{d}}{{dx}}(1)
f(x)=6x+20 f'(x) = 6x + 2 - 0
f(x)=6x+2 f'(x) = 6x + 2

Therefore, the derivative of f(x) = 3x^2 + 2x - 1 is f'(x) = 6x + 2.

c) Using the definition of the derivative, we have:

f(2)=limh0f(2+h)f(2)h f'(2) = \lim_{{h \to 0}} \frac{{f(2 + h) - f(2)}}{h}

Substituting the given function f(x) = 3x^2 + 2x - 1:

f(2)=limh03(2+h)2+2(2+h)1(3(2)2+2(2)1)h f'(2) = \lim_{{h \to 0}} \frac{{3(2 + h)^2 + 2(2 + h) - 1 - (3(2)^2 + 2(2) - 1)}}{h}
f(2)=limh03(4+4h+h2)+4+2h1(12+41)h f'(2) = \lim_{{h \to 0}} \frac{{3(4 + 4h + h^2) + 4 + 2h - 1 - (12 + 4 - 1)}}{h}
f(2)=limh012+12h+3h2+2+2h115h f'(2) = \lim_{{h \to 0}} \frac{{12 + 12h + 3h^2 + 2 + 2h - 1 - 15}}{h}
f(2)=limh03h2+14hh f'(2) = \lim_{{h \to 0}} \frac{{3h^2 + 14h}}{h}
f(2)=limh0(3h+14) f'(2) = \lim_{{h \to 0}} (3h + 14)
f(2)=14 f'(2) = 14

Therefore, the derivative f'(2) using the definition of the derivative is equal to 14.