Answer a-b with the given

Question

Suppose f(x) = cos(3x)b^x and f'(0) = 2 find b

Suppose f(x) = (log_b (x+1))/cos(3x) and f'(0) = 2 find b

Answer 1

Suppose f(x) = cos(3x)b^x

Since this is a product of two terms, we must use the product rule: (u•v)' = u'v + uv'

In this case, u = cos(3x) and v = b^x. That means u' = -3sin(3x) and v' = ln(b)b^x

Putting those together:

(u•v)' = u'v + uv'

f '(x) = (-3sin(3x))(b^x) + (cos(3x))(ln(b)b^x)

f '(x) = b^x[-3sin(3x) + ln(b)cos(3x)]

f '(x) = b^x[ln(b)cos(3x) - 3sin(3x)]

Since f'(0) = 2 we can say:

2 = b⁰[ln(b)cos(3•0) - 3sin(3•0)]

2 = 1[ln(b)•1 - 3•0]

2 = ln(b)

b = e²

Use the same process for the 2nd problem only use the quotient rule (u/v)' = (u'v - uv')/v²