Question

[Calc 1] stuck on how to do this. I know the product and quotient rules, I think if someone can just explain (a) I would be able to do the rest

Answer 1

Part (a)

h(x) = 4*f(x) + 3*g(x)

h ' (x) = d/dx[ 4*f(x) + 3*g(x)]

h ' (x) = d/dx[ 4*f(x) ] + d/dx[ 3*g(x)] ... sum rule

h ' (x) = 4*d/dx[ f(x) ] + 3*d/dx[ g(x)] ... constant multiple rule

h ' (x) = 4*f ' (x) + 3*g ' (x)

h ' (2) = 4*f ' (2) + 3*g ' (2) ... plug in x = 2

h ' (2) = 4*(-2) + 3*(3) ... substitution

h ' (2) = 1

Answer: 1

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Part (b)

h(x) = f(x)*g(x)

h ' (x) = f ' (x)*g(x) + f(x)*g ' (x) ... product rule

h ' (2) = f ' (2)*g(2) + f(2)*g ' (2) ... plug in x = 2

h ' (2) = -2*5 + (-4)*3 ... substitution

h ' (2) = -22

Answer: -22

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Part (c)

h(x) = f(x)/g(x)

h(x) = [f ' (x)*g(x) - g ' (x)*f(x)]/[ (g(x))^2 ] ... quotient rule

h(2) = [f ' (2)*g(2) - g ' (2)*f(2)]/[ (g(2))^2 ] ... plug in x = 2

h(2) = [-2*5 - 3*(-4)]/[ (5)^2 ] ... substitution

h(2) = 2/25

Answer: 2/25

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Part (d)

k(x) = 1 + f(x) ... helps simplify the denominator

k(2) = 1 + f(2) ... plug in x = 2

k(2) = 1 + (-4) ... substitution

k(2) = -3

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k(x) = 1 + f(x)

k ' (x) = 0 + f ' (x)

k ' (x) = f ' (x)

k ' (2) = f ' (2) ... plug in x = 2

k ' (2) = -2

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h(x) = g(x)/(1 + f(x))

h(x) = g(x)/k(x)

h ' (x) = [g ' (x)*k(x) - k ' (x)*g(x)]/[ (k(x))^2 ] ... quotient rule

h ' (2) = [g ' (2)*k(2) - k ' (2)*g(2)]/[ (k(2))^2 ] ... plug in x = 2

h ' (2) = [3*(-3) - (-2)*5]/[ (-3)^2 ] ... substitution

h ' (2) = 1/9

Answer: 1/9