Find Derivative
F(x)= (x-2)(x^2-3x-1)
And I got
F'(x)= (x-2)(3x-3)+ (x^2-3x-1)(1)
yes, correct
but perhaps you better multiply out and collect like terms
for example 3x^2 + x^2 = 4 x^2
Oh okay thanks =D
To find the derivative of the given function F(x) = (x-2)(x^2-3x-1), you can use the product rule, which states that if you have a function in the form of f(x) = g(x) * h(x), then the derivative of f(x) is given by f'(x) = g'(x) * h(x) + g(x) * h'(x).
Let's differentiate each part of the function separately:
1. Differentiating (x-2):
The derivative of x-2 with respect to x is simply 1, since the derivative of x is 1, and the derivative of a constant (in this case, -2) is 0.
So, the derivative of (x-2) is 1.
2. Differentiating (x^2-3x-1):
To differentiate this term, we can use the power rule and linearity property of derivatives.
The power rule states that if you have a term of the form x^n, the derivative is given by n * x^(n-1).
Applying the power rule to the term x^2, we get:
Derivative of x^2 = 2x^(2-1) = 2x
To differentiate the term -3x, we can use the linearity property of derivatives. The derivative of -3x is simply -3 times the derivative of x.
Derivative of -3x = -3 * 1 = -3
For the constant term -1, its derivative is 0, since the derivative of a constant is always 0.
Now, we can rewrite the derivative of (x^2-3x-1) as 2x - 3.
Now, we can use the product rule by multiplying the derivatives of each part:
F'(x) = (x-2)(3x-3) + (x^2-3x-1)(1)
= (3x-3)(x-2) + (x^2-3x-1)(1)
= 3x^2 - 3x - 6x + 6 + x^2 - 3x - 1
= 4x^2 - 12x + 5
Therefore, F'(x) = 4x^2 - 12x + 5.