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Alik [6]
3 years ago
12

For the following exercises, use logarithmic differentiation to find dy/dx.

Mathematics
1 answer:
kondaur [170]3 years ago
8 0

If y=x^{\log_2(x)}, then taking the logarithm of both sides gives

\ln(y) = \ln\left(x^{\log_2(x)}\right) = \log_2(x) \ln(x)

Differentiate both sides with respect to x :

\dfrac{d\ln(y)}{dx} = \dfrac{d\log_2(x)}{dx}\ln(x) + \log_2(x)\dfrac{d\ln(x)}{dx}

\dfrac1y \dfrac{dy}{dx} = \dfrac{d\log_2(x)}{dx}\ln(x) + \dfrac{\log_2(x)}{x}

Now if z=\log_2(x), then 2^z=x. Rewrite

2^z = e^{\ln(2^z)} = e^{\ln(2)z}

Then by the chain rule,

\dfrac{d2^z}{dx} = \dfrac{dx}{dx}

\dfrac{de^{\ln(2)z}}{dx} = 1

e^{\ln(2)z} \ln(2) \dfrac{dz}{dx}= 1

\dfrac{dz}{dx} = \dfrac{1}{e^{\ln(2)z}\ln(2)}

\dfrac{dz}{dx} = \dfrac{1}{2^z \ln(2)}

\dfrac{d\log_2(x)}{dx} = \dfrac{1}{\ln(2)x}

So we have

\dfrac1y \dfrac{dy}{dx} = \dfrac{\ln(x)}{\ln(2)x}+ \dfrac{\log_2(x)}{x}

\dfrac1y \dfrac{dy}{dx} = \dfrac{\log_2(x)}{x}+ \dfrac{\log_2(x)}{x}

\dfrac1y \dfrac{dy}{dx} = \dfrac{2\log_2(x)}{x}

\dfrac1y \dfrac{dy}{dx} = \dfrac{\log_2(x^2)}{x}

\dfrac{dy}{dx} = \dfrac{y\log_2(x^2)}{x}

Replace y :

\dfrac{dy}{dx} = \dfrac{x^{\log_2(x)}\log_2(x^2)}{x}

\dfrac{dy}{dx} = x^{\log_2(x)-1}\log_2(x^2)

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Answer:

1) The linear equation in point and slope form is y - 67 = -4 × (x - 14)

2) The variables are;

a) The number of candies available = y

b) The number of days Jennifer eats the candies =

c) The slope, m = -4

3) Jennifer received 123 pieces of candies on Halloween

Step-by-step explanation:

The given parameters are;

The number of candies Jennifer eats everyday = 4 pieces

The number of days for which Jennifer eats the daily 4 candies = 14

The number of candies left at the end of the 14th day = 67 candies

1) We note that the rate of decrease in the number of candies = 4 candies/day

Therefore, the slope of the linear equation is m = -4

The y-intercept = The initial amount of candies Jennifer has = c = 67 + 14× 4 = 123 candies

The linear equation in point and slope form is given as follows;

y - 67 = -4 × (x - 14)

2) The variables are;

a) The y-value represents the number of candies available on a specific day

b) The x value represents the number of days Jennifer eats the candies'

c) The slope = The rate of decrease in the number of candies per day = -4

3) The number of candies Jennifer receives on Halloween is given by the y-intercept of the straight line equation as follows;

y - 67 = -4 × (x - 14)

y - 67 = -4·x + 56

y = -4·x + 56 + 67 = -4·x + 123

y = -4·x + 123

Comparing the above equation, with the general form of the straight line equation, y = m·x + c, where, the constant term, c = The y-intercept, we have;

The y-intercept of the equation y = -4·x + 123 = 123 = The initial amount of candies Jennifer received on Halloween.

6 0
3 years ago
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