Suppose f(x) and g(x) are functions of x differentiable at x = 4. Given that f(4) = 3, f'(4) = -2, g(4) = 6, g'(4) = -5, find the value of each of the following.

a. Derivative of f(x) * g(x)

My answer: -27

b. Derivative of (f(x))/(g(x))

My answer: 1/12

c. Derivative of (g(x))/(f(x))

My answer: -1/3

d. A tangent line to f(x) at x = 4.

My answer: -2x + 11

looks good to me.

To find the value of each of the given derivatives, we can use the product rule and the quotient rule of derivative.

a. Derivative of f(x) * g(x):

The product rule states that the derivative of the product of two functions is given by the first function times the derivative of the second function, plus the second function times the derivative of the first function.

So, for the derivative of f(x) * g(x), we have:
(f(x) * g(x))' = f(x) * g'(x) + g(x) * f'(x)

Plugging in the values, we get:
(f(x) * g(x))' = f(4) * g'(4) + g(4) * f'(4)
= 3 * (-5) + 6 * (-2)
= -15 - 12
= -27

Therefore, the derivative of f(x) * g(x) is -27.

b. Derivative of (f(x))/(g(x)):

The quotient rule states that the derivative of the quotient of two functions is given by the denominator times the derivative of the numerator, minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

So, for the derivative of (f(x))/(g(x)), we have:
((f(x))/(g(x)))' = (g(x) * f'(x) - f(x) * g'(x))/(g(x))^2

Plugging in the values, we get:
((f(x))/(g(x)))' = (g(4) * f'(4) - f(4) * g'(4))/(g(4))^2
= (6 * (-2) - 3 * (-5))/(6)^2
= (-12 + 15)/(36)
= 3/36
= 1/12

Therefore, the derivative of (f(x))/(g(x)) is 1/12.

c. Derivative of (g(x))/(f(x)):

Similar to the previous case, we can use the quotient rule to find the derivative of (g(x))/(f(x)).

((g(x))/(f(x)))' = (f(x) * g'(x) - g(x) * f'(x))/(f(x))^2

Plugging in the values, we get:
((g(x))/(f(x)))' = (f(4) * g'(4) - g(4) * f'(4))/(f(4))^2
= (3 * (-5) - 6 * (-2))/(3)^2
= (-15 + 12)/(9)
= -3/9
= -1/3

Therefore, the derivative of (g(x))/(f(x)) is -1/3.

d. A tangent line to f(x) at x = 4:

To find the equation of the tangent line to f(x) at x = 4, we need the slope of the tangent line and a point on the line.

The slope of the tangent line at x = 4 can be found using the derivative of f(x). Since f'(4) = -2, the slope of the tangent line is -2.

To find a point on the tangent line, we can use the given point (4, 3). So, the tangent line passes through the point (4, 3).

Using the point-slope form of a line, we can write the equation of the tangent line:

y - y1 = m(x - x1),
where (x1, y1) is the given point and m is the slope.

Plugging in the values, we get:
y - 3 = -2(x - 4)

Simplifying the equation, we have:
y - 3 = -2x + 8
y = -2x + 11

Therefore, the equation of the tangent line to f(x) at x = 4 is y = -2x + 11.