以下の関数の微分を計算し、空欄を埋める問題です。 1. ${(5x^3 + 4x^2 + x + 1)^4}'$

1. 問題の内容

以下の関数の微分を計算し、空欄を埋める問題です。

1. ${(5x^3 + 4x^2 + x + 1)^4}'$

2. ${\left( \left( x - \frac{1}{x} \right)^3 \right)}'$

3. ${\left( \sqrt{x+1} \right)}'$

4. ${\left( \left( -x^3 + 2x^2 + x + 2 \right)^3 \right)}'$

5. ${\left( \sqrt{x^2 + 2x + 2} \right)}'$

2. 解き方の手順

1. ${(5x^3 + 4x^2 + x + 1)^4}' = 4(5x^3 + 4x^2 + x + 1)^3 (15x^2 + 8x + 1)$

$= 4(5x^3 + 4x^2 + x + 1)^3 (15x^2 + 8x + 1)$

よって、(1) = 4, (2) = 3, (3) = 15, (4) = 8, (5) = 1

2. ${\left( \left( x - \frac{1}{x} \right)^3 \right)}' = 3\left( x - \frac{1}{x} \right)^2 \left( 1 + \frac{1}{x^2} \right) = 3\left( x - \frac{1}{x} \right)^2 \left( 1 + \frac{1}{x^2} \right)$

よって、(6) = 3, (7) = 2, (8) = 1, (9) = 1

3. ${\left( \sqrt{x+1} \right)}' = \frac{1}{2\sqrt{x+1}}$

よって、(10) = 2

4. ${\left( \left( -x^3 + 2x^2 + x + 2 \right)^3 \right)}' = 3 \left( -x^3 + 2x^2 + x + 2 \right)^2 \left( -3x^2 + 4x + 1 \right)$

よって、(11) = 3, (12) = 2, (13) = 3, (14) = 4, (15) = 1

5. ${\left( \sqrt{x^2 + 2x + 2} \right)}' = \frac{2x+2}{2\sqrt{x^2 + 2x + 2}} = \frac{x+1}{\sqrt{x^2 + 2x + 2}}$

よって、(16) = 1

3. 最終的な答え

(1) = 4

(2) = 3

(3) = 15

(4) = 8

(5) = 1

(6) = 3

(7) = 2

(8) = 1

(9) = 1

(10) = 2

(11) = 3

(12) = 2

(13) = 3

(14) = 4

(15) = 1

(16) = 1