Find the derivate of the function
f(x) = 2x2 + 23x + 5
What is the value of this derivative when x = 19
F(X) = 2X^2 + 23X + 5.
F'(x) = 4X + 23,
F'(19) = 4*19 + 23 = 99.
To find the derivative of the function f(x) = 2x^2 + 23x + 5, we can use the power rule for differentiation. According to the power rule, if we have a term of the form ax^n, where a is a constant and n is a positive integer, the derivative is given by nax^(n-1).
For the first term, 2x^2, the derivative is:
d/dx (2x^2) = 2 * 2x^(2-1) = 4x^1 = 4x.
For the second term, 23x, the derivative is:
d/dx (23x) = 23 * 1x^(1-1) = 23x^0 = 23.
For the third term, 5, the derivative is zero since it is a constant.
Now, adding up the derivatives of each term, we get:
f'(x) = 4x + 23 + 0
= 4x + 23.
To find the derivative at a specific point, we substitute the given value of x into the derivative function. In this case, we need to find the derivative when x = 19. Plugging x = 19 into f'(x), we get:
f'(19) = 4 * 19 + 23
= 76 + 23
= 99.
Therefore, the value of the derivative of f(x) = 2x^2 + 23x + 5 when x = 19 is 99.