if h(x)= (the antiderivative of) 2x^2 +8x-5 dx
find h'(x)
What is the derivative of the antiderivative?
(what is the square root of the square of a number)?
o so it is 2x^2 +8x-5 dx ... thanks!
wait isn't it 4x+8
cause you still didn't find the derivative of the antiderivative you just reposted the antiderivative
To find h'(x), we need to differentiate the function h(x) with respect to x.
The derivative of a function is the rate at which the function is changing at any given point. In this case, we need to find the derivative of h(x), which will give us the rate of change of h(x) with respect to x.
To differentiate h(x), we can use the power rule of differentiation. The power rule states that the derivative of x^n (where n is any real number) is equal to n*x^(n-1).
Let's differentiate each term of h(x) separately:
1. The derivative of 2x^2:
According to the power rule, the derivative of 2x^2 is equal to 2 * 2x^(2-1) = 4x.
2. The derivative of 8x:
According to the power rule, the derivative of 8x is equal to 8 * x^(1-1) = 8.
3. The derivative of -5:
Since -5 is a constant, its derivative is 0.
Now, we can combine the derivatives of each term to find h'(x):
h'(x) = 4x + 8 + 0
= 4x + 8
Therefore, the derivative of h(x) is h'(x) = 4x + 8.