find the first derivative for
P(x) = a + b (100-x) (x-c)
where a and b is positive constant
To find the first derivative of the function P(x) = a + b(100-x)(x-c), we can use the product rule and apply it step by step.
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) * v(x) + u(x) * v'(x)
In this case, let's set u(x) = a + b(100-x) and v(x) = (x-c). Now, we can find the derivatives of u(x) and v(x) separately.
1. Derivative of u(x):
To find the derivative of u(x), we simply differentiate each term with respect to x since a and b are positive constants. The derivative of a is 0, and the derivative of b(100-x) is -b. So, u'(x) = -b.
2. Derivative of v(x):
To find the derivative of v(x), we differentiate it with respect to x. The derivative of x is 1, and the derivative of -c is 0. So, v'(x) = 1.
Now, we can apply the product rule:
(d/dx)(u(x) * v(x)) = u'(x) * v(x) + u(x) * v'(x)
= (-b) * (x-c) + (a + b(100-x)) * 1
= -bx + bc + a + b(100-x)
= a + bx - bc - b*x + b*100 - b*x
= a - bc + 2bx - 2bx
= a - bc
So, the first derivative of P(x) = a + b(100-x)(x-c) is P'(x) = a - bc.
Therefore, the first derivative of P(x) is a - bc.