Suppose that f is a differentiable function, defined everywhere, such that f\'(x

Question

Suppose that f is a differentiable function, defined everywhere, such that f'(x) = cos^2(x), and f(4) = pi/4. Find the equation of the tangent line to the curve. y = f(ln(x)) + x^2 at x = e^4. You have a debt of $700 on a credit card with a monthly interest rate of 10%. After t months, your debt is d(t) = 700 (1.1)^t dollars. Write d(t) in the form d(t) = Ce^kt for some appropriate constants C and k. How much do you owe after one year What is the rate of change of your debt after one year (Remember to include units.) How many months will it be until you owe one million dollars Since the derivative of ln(x) is 1/x, we know that, by the definition of the derivative, lim_A rightarrow 0 ln(2 +h) - ln(2) / h = 1/2. Use this fact to show carefully, using the definition of the derivative, that the function f(x) = |ln(x) - ln(2)| is not differentiable at x = 2.

Explanation / Answer

1.

first find y'
y' = f'(ln(x))/x + 2x

the slope at x = e^4 is y'(e^4)
y'(e^4) = f'(ln(e^4))/e^4 + 2e^4
= f'(4)/e^4 + 2e^4

we're given f'(x) = cos^2(x), thus f'(4) = cos^2(4)
y'(e^4) = cos^2(4)/e^4 + 2e^4 <--- this is the slope of the tangent line.

y has to go through the point (e^4, y(e^4)). So we need to find the value of y(e^4).

y(e^4) = f(ln(e^4)) + (e^4)^2
= f(4) + e^8

and we're given f(4) = pi/4, so
y(e^4) = pi/4 + e^8

using point slope form, we have:
y - (pi/4 + e^8) = [cos^2(4)/e^4 + 2e^4] (x - e^4)

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