Differentiate the function.
g(x)= x^2(1-2x)
g'(x)=
i tried doing
2x(1-2x)
2x-4x^2
never mind I got it but could someone just help me with this one.
A(s)= -11/s^6
A(s) = -11 * s^-6
so,
A' = 66*s^-7 = 66/s^7
Thank you!!
To differentiate the function g(x) = x^2(1-2x), we will use the product rule.
The product rule states that if we have two functions u(x) and v(x), then their derivative is given by:
(uv)' = u'v + uv'
So, let's differentiate the function step by step:
Step 1: Identify u(x) and v(x)
In this case, u(x) = x^2 and v(x) = (1-2x).
Step 2: Find the derivatives of u(x) and v(x)
The derivative of u(x) = x^2 is found using the power rule:
u'(x) = 2x.
The derivative of v(x) = (1-2x) is found using the constant rule:
v'(x) = -2.
Step 3: Apply the product rule
Using the product rule, the derivative of g(x) is given by:
g'(x) = u'v + uv'
= (2x)(1-2x) + (x^2)(-2)
= 2x - 4x^2 - 2x^2
= -4x^2 + 2x - 2x^2
= -6x^2 + 2x
So, the derivative of g(x) is g'(x) = -6x^2 + 2x.