The derivatives for this problem are given as follows:
a) -3.
b) -1.
c) 1.
d) 0.
What is the product rule for a derivative?
The product rule for a derivative is given as follows:
[f(x)g(x)]' = f'(x)g(x) + g'(x)f(x).
Hence, at x = 6, we have that:
[f(x)g(x)]'(6) = f'(6)g(6) + g'(6)f(6).
Replacing the values given in this problem, we have that the answer for item a is:
[f(x)g(x)]'(6) = f'(6)g(6) + g'(6)f(6) = 2(-1) - 1(1) = -2 - 1 = -3.
What is the quotient rule for a derivative?
The product rule for a derivative is given as follows:
[tex]\left(\frac{f(x)}{g(x)}\right)^{\prime} = \frac{f^{\prime}(x)g(x) - g^{\prime}(x)f(x)}{g(x)^2}[/tex]
Hence, at x = 6, we have that:
[tex]\left(\frac{f(x)}{g(x)}\right)^{\prime}(6) = \frac{f^{\prime}(6)g(6) - g^{\prime}(6)f(6)}{g(6)^2}[/tex]
Then the derivative in item b is:
[2(-1) - (-1)(1)]/[(-1)^2] = -1/1 = -1.
What is the derivative for the square root of a function?
Applying the chain rule, the derivative is given by:
[tex][\sqrt{f(x)}]^{\prime} = \frac{1}{2\sqrt{f(x)}}f^{\prime}(x)[/tex]
Hence, at x = 6:
[tex][\sqrt{f(x)}]^{\prime}(6) = \frac{1}{2\sqrt{f(6)}}f^{\prime}(6) = 2/2 = 1[/tex]
Hence the derivative in item c is of 1.
What is the derivative of a constant?
The derivative of a constant is of 0. In item d, the multiplication of f(6) by g'(6) results in a constant, hence the derivative is of 0.
More can be learned about derivative rules at https://brainly.com/question/25081524
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