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Chapter 3: Problem 28

Find the derivatives of the following functions. $$P=\frac{40}{1+2^{-t}}$$

Short Answer

Expert verified
Answer: The derivative of the function with respect to t is $$\frac{dP}{dt} = \frac{40\ln 2 *2^{-t}}{(1+2^{-t})^2}$$.

Step by step solution

01

Identify the quotient rule formula

The quotient rule formula is used to find the derivative of a composite function that is a quotient of two functions. The formula is given by: $$\frac{d}{dt}\left(\frac{f(t)}{g(t)}\right) = \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2}$$ where f(t) and g(t) are the numerator and denominator functions, and f'(t) and g'(t) are their derivatives with respect to t.
02

Identify the functions f(t) and g(t)

In this problem, the given function is: $$P=\frac{40}{1+2^{-t}}$$ From this, we can identify the numerator and denominator functions as: $$f(t) = 40$$ and $$g(t) = 1 + 2^{-t}$$
03

Find the derivatives of f(t) and g(t)

The derivative of a constant is 0, so: $$f'(t) = 0$$ To find the derivative of g(t), we will need to apply the chain rule, which states: $$\frac{d}{dt}(h(u(t))) = h'(u(t)) * u'(t)$$ In this case, our inner function u(t) is -t, and our outer function h(u) is 2^u. So: $$u'(t) = -1$$ $$h'(u) = 2^u \ln 2$$ Then the derivative of g(t) is: $$g'(t) = h'(u(t)) * u'(t) = 2^{-t} \ln 2 * (-1) = -2^{-t} \ln 2$$
04

Apply the quotient rule

Now that we have the derivatives of f(t) and g(t), we can apply the quotient rule to find the derivative of P: $$\frac{dP}{dt} = \frac{f'(t)g(t) - f(t)g'(t)}{(g(t))^2}$$ Substitute our values: $$\frac{dP}{dt} = \frac{0*(1+2^{-t}) - 40*(-2^{-t} \ln 2)}{(1+2^{-t})^2}$$
05

Simplify the expression

Simplify the expression to find the final form of the derivative: $$\frac{dP}{dt} = \frac{40*(-2^{-t} \ln 2)}{(1+2^{-t})^2}$$ Thus, the derivative of the given function is: $$\frac{dP}{dt} = \frac{40\ln 2 *2^{-t}}{(1+2^{-t})^2}$$

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