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) = -4, 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 (not sure)

a) let y = f(x) * g(x)

then y' = f(x)* g '(x) + g(x) * f '(x)
given:
f(4) = 3
f '(4) = -2
g(4) = 6
g '(4) = -4
so
y' = 3(-4) + 6(-2) = -24

how did you get -27 ?

do the others the same way

I'm sorry, g'(x) = -5.

To find the value of each expression, let's use the rules of differentiation.

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

The product rule states that the derivative of the product of two functions is equal to the derivative of the first function multiplied by the second function, plus the first function multiplied by the derivative of the second function.

So, the derivative of f(x) * g(x) is:
(f(x) * g(x))' = f'(x) * g(x) + f(x) * g'(x)

Substituting the values we have:
(f(x) * g(x))' = f'(4) * g(4) + f(4) * g'(4)
(f(x) * g(x))' = (-2) * 6 + 3 * (-4)
(f(x) * g(x))' = -12 - 12
(f(x) * g(x))' = -24

Therefore, the derivative of f(x) * g(x) at x = 4 is -24

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

To find the derivative of the quotient of two functions, we can use the quotient rule. The quotient rule states that the derivative of the quotient of two functions is equal to (denominator * derivative of the numerator - numerator * derivative of the denominator) divided by (denominator)^2.

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

Substituting the values we have:
((f(x))/(g(x)))' = ((6 * (-2)) - (3 * (-4)))/(6)^2
((f(x))/(g(x)))' = (-12 + 12)/(36)
((f(x))/(g(x)))' = 0/36
((f(x))/(g(x)))' = 0

Therefore, the derivative of (f(x))/(g(x)) at x = 4 is 0.

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

Just like in part b, we can use the quotient rule to find the derivative. The derivative of (g(x))/(f(x)) is:
((g(x))/(f(x)))' = ((f(x) * g'(x)) - (g(x) * f'(x)))/(f(x))^2

Substituting the values we have:
((g(x))/(f(x)))' = ((3 * (-4)) - (6 * (-2)))/(3)^2
((g(x))/(f(x)))' = (-12 + 12)/(9)
((g(x))/(f(x)))' = 0/9
((g(x))/(f(x)))' = 0

Therefore, the derivative of (g(x))/(f(x)) at x = 4 is 0.

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

To find the equation of a tangent line, we need the slope of the tangent line, which is the value of the derivative at the given point. In this case, the derivative of f(x) is f'(x) = -2.

Using the point-slope form of a linear equation, the equation of the tangent line is:
y - y1 = m(x - x1)

Substituting the known values:
y - f(4) = f'(4)(x - 4)
y - 3 = -2(x - 4)
y - 3 = -2x + 8
y = -2x + 11

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