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Given that $y = x^4 + x^3 + 3$, find \( \frac{dy}{dx} \). - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 2

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Given-that-$y-=-x^4-+-x^3-+-3$,-find-\(-\frac{dy}{dx}-\).-Edexcel-A-Level Maths Pure-Question 4-2010-Paper 2.png

Given that $y = x^4 + x^3 + 3$, find \( \frac{dy}{dx} \).

Worked Solution & Example Answer:Given that $y = x^4 + x^3 + 3$, find \( \frac{dy}{dx} \). - Edexcel - A-Level Maths Pure - Question 4 - 2010 - Paper 2

Step 1

Step 1: Find \( \frac{dy}{dx} \)

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Answer

To find ( \frac{dy}{dx} ), we need to differentiate the function (y = x^4 + x^3 + 3). Using the power rule of differentiation, we differentiate each term:

  1. The derivative of (x^4) is (4x^3).
  2. The derivative of (x^3) is (3x^2).
  3. The derivative of a constant (3) is 0.

Thus, we have:

dydx=4x3+3x2+0=4x3+3x2.\frac{dy}{dx} = 4x^3 + 3x^2 + 0 = 4x^3 + 3x^2.

Step 2

Step 2: Evaluate at critical points

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Answer

Next, if we set ( \frac{dy}{dx} = 0 ) to find the critical points, we get:

4x3+3x2=04x^3 + 3x^2 = 0.

Factoring out the common term:

x2(4x+3)=0x^2(4x + 3) = 0.

This gives us (x = 0) or (4x + 3 = 0), which leads to (x = -\frac{3}{4}). These points can be further investigated for maxima, minima, or points of inflection.

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