Find the derivative of the function.
f(x) = (x+1)(4x^2-8x+7)
use the product rule:
(first times der of second) +
(second times der of first)
(x+1)(4x^2-8x+7)
u = x + 1
v = 4x^2 - 8x + 7
Use the product rule,
d/dx (uv) = u dv/dx + v du/dx
d/dx (uv) = (x+1)(8x-8)+(4x^2-8x+7)(1)
You can take it from here
To find the derivative of the given function, we will use the product rule and the power rule of differentiation.
The product rule states that if we have two functions u(x) and v(x), then the derivative of their product is given by:
(d/dx)(u(x)v(x)) = u(x)(d/dx)v(x) + v(x)(d/dx)u(x)
In this case, u(x) = (x+1) and v(x) = (4x^2-8x+7).
First, let's find the derivative of u(x):
(d/dx)(x+1) = 1 (The derivative of x is 1, and the derivative of a constant 1 is 0.)
Now, let's find the derivative of v(x):
(d/dx)(4x^2-8x+7) = 8x - 8 (Using the power rule, we bring down the exponent and multiply by the coefficient, then subtracting 1 from the exponent.)
Using the product rule, the derivative of f(x) = (x+1)(4x^2-8x+7) is given by:
= u(x)(d/dx)v(x) + v(x)(d/dx)u(x)
= (x+1)(8x - 8) + (4x^2-8x+7)(1)
= 8x(x+1) - 8(x+1) + 4x^2-8x+7
= 8x^2 + 8x - 8x - 8 + 4x^2 - 8x + 7
= 12x^2 - 1
So, the derivative of f(x) = (x+1)(4x^2-8x+7) is 12x^2 - 1.