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creativ13 [48]
2 years ago
5

How to differentiate ?

Mathematics
1 answer:
Bas_tet [7]2 years ago
7 0

Use the power, product, and chain rules:

y = x^2 (3x-1)^3

• product rule

\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{\mathrm d(x^2)}{\mathrm dx}\times(3x-1)^3 + x^2\times\dfrac{\mathrm d(3x-1)^3}{\mathrm dx}

• power rule for the first term, and power/chain rules for the second term:

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(x-1)^2\times\dfrac{\mathrm d(3x-1)}{\mathrm dx}

• power rule

\dfrac{\mathrm dy}{\mathrm dx} = 2x\times(3x-1)^3 + x^2\times3(3x-1)^2\times3

Now simplify.

\dfrac{\mathrm dy}{\mathrm dx} = 2x(3x-1)^3 + 9x^2(3x-1)^2 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2 \times (2(3x-1) + 9x) \\\\ \boxed{\dfrac{\mathrm dy}{\mathrm dx} = x(3x-1)^2(15x-2)}

You could also use logarithmic differentiation, which involves taking logarithms of both sides and differentiating with the chain rule.

On the right side, the logarithm of a product can be expanded as a sum of logarithms. Then use other properties of logarithms to simplify

\ln(y) = \ln\left(x^2(3x-1)^3\right) \\\\ \ln(y) =  \ln\left(x^2\right) + \ln\left((3x-1)^3\right) \\\\ \ln(y) = 2\ln(x) + 3\ln(3x-1)

Differentiate both sides and you end up with the same derivative:

\dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac2x + \dfrac9{3x-1} \\\\ \dfrac1y\dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \\\\ \dfrac{\mathrm dy}{\mathrm dx} = \dfrac{15x-2}{x(3x-1)} \times x^2(3x-1)^3 \\\\ \dfrac{\mathrm dy}{\mathrm dx} = x(15x-2)(3x-1)^2

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HELP PLSS
atroni [7]

The balloon reaches a height of 7 feet at 0.1 seconds and 2.13 seconds

<h3>How to determine the time the balloon is at the height?</h3>

The equation of the function is given as

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The above equation is a quadratic equation

When the balloon is at a height of 7 feet, we have

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So, we have the following equations

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Read more about quadratic equation at

brainly.com/question/15709421

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= 1292 – 625

= 667

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