Question

attempt 1: 10 attempts remaining. calculate the derivative of the function. ( f(x) = (8x - 1)^2 ) ( f(x) = ) submit answer next item

Explanation:

Step1: Identify the function type

The function \( f(x) = (8x - 1)^2 \) is a composite function. We can use the chain rule or expand the function first. Let's expand it for simplicity.
Expanding \( (8x - 1)^2 \) using the formula \( (a - b)^2 = a^2 - 2ab + b^2 \), where \( a = 8x \) and \( b = 1 \).
So, \( f(x) = (8x)^2 - 2\times(8x)\times1 + 1^2 = 64x^2 - 16x + 1 \).

Step2: Differentiate term by term

Using the power rule for differentiation, which states that if \( y = x^n \), then \( y' = nx^{n - 1} \).

• For the term \( 64x^2 \), the derivative is \( 64\times2x^{2 - 1} = 128x \).
• For the term \( -16x \), the derivative is \( -16\times1x^{1 - 1} = -16 \).
• For the constant term \( 1 \), the derivative is \( 0 \) (since the derivative of a constant is 0).

Step3: Combine the derivatives

Adding the derivatives of each term together: \( f'(x) = 128x - 16 \).

We can also use the chain rule to verify. The chain rule states that if \( f(x) = g(h(x)) \), then \( f'(x) = g'(h(x))\times h'(x) \). Let \( g(u) = u^2 \) and \( h(x) = 8x - 1 \). Then \( g'(u) = 2u \) and \( h'(x) = 8 \). Substituting back, we get \( f'(x) = 2(8x - 1)\times8 = 16(8x - 1) = 128x - 16 \), which matches the result from expanding.

Answer:

\( 128x - 16 \)