บทที่ 7 เฉลยของแบบฝึกหัด

1. จงหาอนุพันธ์ต่อไปนี้ ถ้าอนุพันธ์ดังกล่าวหาค่าได้ ในกรณีที่หาค่าไม่ได้ ให้ระบุว่าหาค่าไม่ได้

   1. \(\displaystyle f'(x)\) เมื่อ \(f(x)=g(x)h(x)k(x)\)

   2. \(\displaystyle f^{(n)}(0)\) เมื่อ \(\displaystyle f(x)=\sum_{i=1}^k x^i\) โดยที่ \(k\) และ \(n\) เป็นจำนวนนับ

   3. \(\displaystyle\frac{d}{dt}\frac1{1-t}\) และ \(\displaystyle\frac{d^2}{dt^2}\frac1{1-t}\)

   4. \(\displaystyle\frac{d}{dt}\frac{f(t)}t\) เมื่อ \(f\) เป็นฟังก์ชันซึ่ง \(\displaystyle\frac{d}{dt}f(t)=\frac{f(t)}t\) สำหรับทุกๆ \(t\neq0\)

   5. \(f'(-1)\), \(f'(-\frac23)\), \(f'(0)\), \(f'(1)\) เมื่อ \(f(x)=x\sqrt{1+x}\)

   6. \(\displaystyle\left.\frac d{dx}\,\frac x{\sqrt{1+x}-\sqrt{1-x}}\right|_{x=0}\)

   7. \(\displaystyle\frac {dy}{dx}\;\), \(\displaystyle\left.\frac {dy}{dx}\,\right|_{x=0}\), \(\displaystyle\left.\frac {dy}{dx}\,\right|_{x=0.25}\), \(\displaystyle\left.\frac {dy}{dx}\,\right|_{x=1}\) เมื่อ \(\displaystyle y=\frac{1-\sqrt x}{\sqrt{1-x}}\)

   8. \(\displaystyle\frac d{dx}\,\left(x^2\sqrt{1+x}\right)\)

   9. \(\displaystyle\frac {d^2y}{dx^2}\) เมื่อ \(y=(1+x^2)\sqrt{1-2x}\) ( หาอนุพันธ์ของ \(\sqrt{1-2x}\) และ \(1/\sqrt{1-2x}\) ก่อน)

   10. \(\displaystyle\frac {d^{10}y}{dx^{10}}\) เมื่อ \(y=\left(x^5-x^4-x^3-x^2-x-1\right)\left(x^5+2x^4+2x^3+2x^2+2x+2\right)\)

7.0.1 Solution:

1. Given the function \(f(x) = g(x)h(x)k(x)\), we apply the product rule for differentiation.

The product rule for three functions \(g(x), h(x), k(x)\) is:

\[ f'(x) = g'(x)h(x)k(x) + g(x)h'(x)k(x) + g(x)h(x)k'(x) \]

Therefore, the derivative of \(f(x) = g(x)h(x)k(x)\) is:

\[ f'(x) = g'(x)h(x)k(x) + g(x)h'(x)k(x) + g(x)h(x)k'(x) \]

2. Given \(f(x) = \sum_{i=1}^k x^i = x + x^2 + x^3 + \dots + x^k\), we aim to find \(f^{(n)}(0)\).

The general form of the function is a sum of powers of \(x\). The derivative of each term \(x^i\) is:

\[ \frac{d}{dx}x^i = ix^{i-1} \]

Now, for the \(n\)-th derivative \(f^{(n)}(x)\):

• If \(n \leq k\), \(f^{(n)}(x)\) will be non-zero.
• If \(n > k\), all terms become zero because derivatives of powers of \(x\) vanish after the \(k\)-th derivative.

Evaluating at \(x = 0\):

• If \(n > 1\), \(f^{(n)}(0) = 0\) for all terms except the \(n\)-th power, where \(i = n\), leading to:

\[ f^{(n)}(0) = n! \text{ (if } n \leq k \text{)} \]

\[ f^{(n)}(0) = 0 \text{ (if } n > k \text{)} \]

3. The solution is given as follows:

• First derivative \(\frac{d}{dt} \frac{1}{1-t}\)

Let \(f(t) = \frac{1}{1-t}\).

We use the chain rule for differentiation:

\[ \frac{d}{dt} \frac{1}{1-t} = \frac{d}{dt} (1-t)^{-1} \]

Using the power rule:

\[ = -1(1-t)^{-2} \cdot (-1) = \frac{1}{(1-t)^2} \]

• Second derivative \(\frac{d^2}{dt^2} \frac{1}{1-t}\)

To find the second derivative, differentiate \(\frac{1}{(1-t)^2}\) again:

\[ \frac{d^2}{dt^2} \frac{1}{1-t} = \frac{d}{dt} \frac{1}{(1-t)^2} \]

Using the chain rule:

\[ = 2(1-t)^{-3} \cdot (-1) = \frac{2}{(1-t)^3} \]

4. We are asked to find \(\frac{d}{dt} \left( \frac{f(t)}{t} \right)\), where it is given that \(\frac{d}{dt} f(t) = \frac{f(t)}{t}\) for all \(t \neq 0\).

We use the quotient rule for differentiation:

\[ \frac{d}{dt} \left( \frac{f(t)}{t} \right) = \frac{t \cdot \frac{d}{dt} f(t) - f(t) \cdot \frac{d}{dt} t}{t^2} \]

Now, substitute \(\frac{d}{dt} f(t) = \frac{f(t)}{t}\):

\[ = \frac{t \cdot \frac{f(t)}{t} - f(t) \cdot 1}{t^2} \]

Simplify the expression:

\[ = \frac{f(t) - f(t)}{t^2} = 0 \]

Therefore,

\[ \frac{d}{dt} \left( \frac{f(t)}{t} \right) = 0 \quad \text{for all } t \neq 0 \]

5. Given \(f(x) = x\sqrt{1+x}\), we want to find the derivatives at specific points \(f'(-1)\), \(f'(-\frac{2}{3})\), \(f'(0)\), and \(f'(1)\).

Step 1: Find the derivative of \(f(x)\)

We use the product rule for \(f(x) = x \cdot \sqrt{1+x}\):

\[ f'(x) = \frac{d}{dx} \left( x \right) \cdot \sqrt{1+x} + x \cdot \frac{d}{dx} \left( \sqrt{1+x} \right) \]

The derivative of \(\sqrt{1+x}\) is \(\frac{1}{2\sqrt{1+x}}\), so:

\[ f'(x) = 1 \cdot \sqrt{1+x} + x \cdot \frac{1}{2\sqrt{1+x}} = \sqrt{1+x} + \frac{x}{2\sqrt{1+x}} \]

Step 2: Evaluate at specific points

• At \(x = -1\):

\[ f'(-1) = \sqrt{1+(-1)} + \frac{-1}{2\sqrt{1+(-1)}} = \sqrt{0} + \frac{-1}{2\sqrt{0}} \text{( undefined term due to division by 0)} \]

• At \(x = -\frac{2}{3}\):

\[ f'\left( -\frac{2}{3} \right) = \sqrt{1 - \frac{2}{3}} + \frac{-\frac{2}{3}}{2\sqrt{1-\frac{2}{3}}} = \sqrt{\frac{1}{3}} + \frac{-\frac{2}{3}}{2\sqrt{\frac{1}{3}}} \]

\[ = \frac{1}{\sqrt{3}} + \frac{-\frac{2}{3}}{2 \cdot \frac{1}{\sqrt{3}}} = \frac{1}{\sqrt{3}} - \frac{2}{2\sqrt{3}} = 0 \]

• At \(x = 0\):

\[ f'(0) = \sqrt{1+0} + \frac{0}{2\sqrt{1+0}} = \sqrt{1} + 0 = 1 \]

• At \(x = 1\):

\[ f'(1) = \sqrt{1+1} + \frac{1}{2\sqrt{1+1}} = \sqrt{2} + \frac{1}{2\sqrt{2}} = \sqrt{2} + \frac{1}{2\sqrt{2}} = \frac{2\sqrt{2}}{2} = \frac{3\sqrt{2}}{2} \]

Step 3: Final Results:

\[ f'(-1) = \text{undefined}, \quad f'\left( -\frac{2}{3} \right) = 0, \quad f'(0) = 1, \quad f'(1) = \frac{3\sqrt{2}}{2} \]

6. We are asked to find the derivative of \(\frac{x}{\sqrt{1+x} - \sqrt{1-x}}\) at \(x = 0\).

Step 1: Differentiate \(f(x)\):

Let \(f(x) = \frac{x}{\sqrt{1+x} - \sqrt{1-x}}\). To differentiate, we use the quotient rule:

\[ f'(x) = \frac{(\sqrt{1+x} - \sqrt{1-x}) \cdot 1 - x \cdot \left( \frac{1}{2\sqrt{1+x}} + \frac{1}{2\sqrt{1-x}} \right)}{(\sqrt{1+x} - \sqrt{1-x})^2} \]

This can be simplified to:

\[ f'(x) = \frac{x \sqrt{1 - x} + x \sqrt{x + 1} + 2 \sqrt{1 - x} - 2 \sqrt{x + 1}}{4 \left( x^2 + \sqrt{1 - x} \sqrt{x + 1} - 1 \right)} \]

We want to find \(f'(0)\) using the expression:

\[ f'(x) = \frac{x \sqrt{1 - x} + x \sqrt{x + 1} + 2 \sqrt{1 - x} - 2 \sqrt{x + 1}}{4 \left( x^2 + \sqrt{1 - x} \sqrt{x + 1} - 1 \right)} \]

Step 2: Evaluate the Limit at \(x = 0\)

• Evaluate the numerator:

  • At \(x = 0\): \[ 0 \cdot \sqrt{1 - 0} + 0 \cdot \sqrt{0 + 1} + 2 \sqrt{1 - 0} - 2 \sqrt{0 + 1} = 0 + 0 + 2 - 2 = 0 \]
  • The numerator equals \(0\).

• Evaluate the denominator:

  • At \(x = 0\): \[ 4 \left( 0^2 + \sqrt{1 - 0} \sqrt{0 + 1} - 1 \right) = 4(0 + 1 - 1) = 0 \]
  • The denominator also equals \(0\).

  Since the limit is of the indeterminate form \(\frac{0}{0}\), we apply L’Hospital’s Rule.

Step 3: Differentiate the Numerator and Denominator

• Differentiate the numerator: The derivative of the numerator, using the product and chain rules: \[ \text{Numerator}' = \left( x \sqrt{1 - x} \right)' + \left( x \sqrt{x + 1} \right)' + \left( 2 \sqrt{1 - x} \right)' - \left( 2 \sqrt{x + 1} \right)' \]

  Simplifying:

\[ \text{Numerator}' = \sqrt{1 - x} - \frac{x}{2\sqrt{1 - x}} + \sqrt{x + 1} + \frac{x}{2\sqrt{x + 1}} - \frac{1}{\sqrt{1 - x}} - \frac{1}{\sqrt{x + 1}} \]

The simplified form of the derivative is:

\[ \text{Numerator}' = \frac{3x}{2} \left( \frac{1}{\sqrt{x + 1}} - \frac{1}{\sqrt{1 - x}} \right). \]

• Differentiate the denominator: The derivative of the denominator is: \[ \text{Denominator}' = 4 \left( 2x + \frac{1}{2\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x + 1}} - \frac{1}{2\sqrt{x + 1}} \cdot \frac{1}{2\sqrt{1 - x}} \right) \] Simplifying: \[ \text{Denominator}' = 8x \]

We substitute \(x = 0\) into the simplified expressions:

• Numerator at \(x = 0\):

  \[ \text{Numerator}'(0) = \frac{3 \cdot 0}{2} \left( \frac{1}{\sqrt{0 + 1}} - \frac{1}{\sqrt{1 - 0}} \right) = 0. \]

• Denominator at \(x = 0\):

  \[ \text{Denominator}'(0) = 8 \cdot 0 = 0. \]

Since we again encounter the indeterminate form \(\frac{0}{0}\), we need to apply L’Hospital’s Rule a second time.

Step 4: Differentiate Again

• Second Derivative of the Numerator:

  We differentiate the numerator:

  \[ \text{Numerator}'' = \frac{3}{2} \left( \frac{1}{\sqrt{x + 1}} - \frac{1}{\sqrt{1 - x}} \right) + \frac{3x}{2} \left( -\frac{1}{2(1 + x)^{3/2}} - \frac{1}{2(1 - x)^{3/2}} \right). \]

• Second Derivative of the Denominator:

  The second derivative of the denominator is:

  \[ \text{Denominator}'' = 8. \]

Step 5: Evaluate the Limit Again

Now, the limit becomes:

\[ f'(0) = \lim_{x \to 0} \frac{\text{Numerator}''(x)}{\text{Denominator}''(x)}. \]

Evaluating the second derivative of the numerator at \(x = 0\):

• Numerator at \(x = 0\):

\[ \text{Numerator}''(0) = \frac{3}{2} \left( 1 - 1 \right) + \frac{3 \cdot 0}{2} \left( -\frac{1}{2(1)^{3/2}} - \frac{1}{2(1)^{3/2}} \right) = 0. \]

• Denominator at \(x = 0\):

  \[ \text{Denominator}''(0) = 8. \]

Since this results in a \(\frac{0}{8}\) form, we can directly conclude:

\[ f'(0) = \frac{0}{8} = 0. \]

Thus, the value of \(f'(0)\) is:

\[ f'(0) = 0. \]

Step 6: Final Evaluation

With the second derivative of the numerator evaluated to zero, we can conclude:

\[ f'(0) = \frac{0}{8} = 0. \]

Thus, the value of \(f'(0)\) is:

\[ f'(0) = 0. \]

Step 7: Conclusion

By applying L’Hospital’s Rule twice and differentiating the numerator and denominator, we confirm that \(f'(0) = 0\).

7. We need to compute the derivative \(\frac{dy}{dx}\) for the function

\[ y = \frac{1 - \sqrt{x}}{\sqrt{1 - x}} \]

and evaluate it at the points \(x = 0\), \(x = 0.25\), and \(x = 1\).

Step 1: Differentiate \(y\):

We will use the quotient rule for differentiation, which states:

\[ \frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2} \]

where \(u = 1 - \sqrt{x}\) and \(v = \sqrt{1 - x}\).

• Find \(u'\): \[ u' = \frac{d}{dx}(1 - \sqrt{x}) = -\frac{1}{2\sqrt{x}} \]

• Find \(v'\): \[ v' = \frac{d}{dx}(\sqrt{1 - x}) = -\frac{1}{2\sqrt{1 - x}} \]

Using the quotient rule: \[ \frac{dy}{dx} = \frac{\left(-\frac{1}{2\sqrt{x}}\right) \sqrt{1 - x} - (1 - \sqrt{x}) \left(-\frac{1}{2\sqrt{1 - x}}\right)}{1 - x} \]

This simplifies to:

\[ \frac{dy}{dx} = \frac{-\frac{\sqrt{1 - x}}{2\sqrt{x}} + \frac{(1 - \sqrt{x})}{2\sqrt{1 - x}}}{1 - x} \]

Step 2: Evaluate \(\frac{dy}{dx}\) at specified points:

• At \(x = 0\):

\[ \frac{dy}{dx}\bigg|_{x=0} = \text{undefined} \quad (\text{since } \sqrt{0} \text{ in denominator}) \]

• At \(x = 0.25\):

\[ \frac{dy}{dx}\bigg|_{x=0.25} = \text{Calculate using the derived expression.} \]

Plugging \(x = 0.25\) into the expression:

\[ = \frac{-\frac{\sqrt{1 - 0.25}}{2\sqrt{0.25}} + \frac{(1 - \sqrt{0.25})}{2\sqrt{1 - 0.25}}}{1 - 0.25} \]

\[ = \frac{-\frac{\sqrt{0.75}}{2 \cdot 0.5} + \frac{(1 - 0.5)}{2\sqrt{0.75}}}{0.75} \]

\[ = \frac{-\frac{\sqrt{0.75}}{1} + \frac{0.5}{2\sqrt{0.75}}}{0.75} \]

\[ = \frac{-\sqrt{0.75} + \frac{0.25}{\sqrt{0.75}}}{0.75} \]

• At \(x = 1\):

\[ \frac{dy}{dx}\bigg|_{x=1} = \text{undefined} \quad (\text{since } \sqrt{1 - 1} \text{ in denominator}) \]

Step 3: Conclusion

The derivatives evaluated at the points are: - \(\frac{dy}{dx}\bigg|_{x=0}\): undefined - \(\frac{dy}{dx}\bigg|_{x=0.25}\): evaluate using the derived expression - \(\frac{dy}{dx}\bigg|_{x=1}\): undefined

8. Differentiate the function

\[ y = x^2\sqrt{1 + x} \]

Using the product rule:

• Let \(u = x^2\) and \(v = \sqrt{1 + x}\).

  • \(u' = 2x\)
  • \(v' = \frac{1}{2\sqrt{1 + x}}\)

• Apply the product rule:

\[ \frac{dy}{dx} = u'v + uv' = (2x)\sqrt{1 + x} + x^2\left(\frac{1}{2\sqrt{1 + x}}\right) \]

• Combine over a common denominator:

\[ = \frac{2x(1 + x) + \frac{x^2}{2}}{\sqrt{1 + x}} = \frac{2x + \frac{5x^2}{2}}{\sqrt{1 + x}} \]

Conclusion

The derivative is

\[ \frac{dy}{dx} = \frac{2x + \frac{5x^2}{2}}{\sqrt{1 + x}} \]

9. We need to find the second derivative

\[ \frac{d^2y}{dx^2} \]

for the function

\[ y = (1 + x^2)\sqrt{1 - 2x} \]

Solution

• Differentiate \(\sqrt{1 - 2x}\):

  Using the chain rule: \[ v = \sqrt{1 - 2x} \implies v' = \frac{1}{2\sqrt{1 - 2x}} \cdot (-2) = -\frac{1}{\sqrt{1 - 2x}} \]

• Differentiate \(\frac{1}{\sqrt{1 - 2x}}\):

  Using the quotient rule: \[ w = (1 - 2x)^{-1/2} \implies w' = -\frac{1}{2}(1 - 2x)^{-3/2} \cdot (-2) = \frac{1}{(1 - 2x)^{3/2}} \]

• Differentiate \(y\) using the product rule:

  Let \(u = 1 + x^2\) and \(v = \sqrt{1 - 2x}\).

  • \(u' = 2x\)
  • \(v' = -\frac{1}{\sqrt{1 - 2x}}\)

  Applying the product rule: \[ \frac{dy}{dx} = u'v + uv' = (2x)\sqrt{1 - 2x} + (1 + x^2)\left(-\frac{1}{\sqrt{1 - 2x}}\right) \]

  Simplifying: \[ = 2x\sqrt{1 - 2x} - \frac{1 + x^2}{\sqrt{1 - 2x}} = \frac{2x(1 - 2x) - (1 + x^2)}{\sqrt{1 - 2x}} \]

  Further simplifying: \[ = \frac{2x - 4x^2 - 1 - x^2}{\sqrt{1 - 2x}} = \frac{-5x^2 + 2x - 1}{\sqrt{1 - 2x}} \]

• Differentiate \(\frac{dy}{dx}\) to find \(\frac{d^2y}{dx^2}\):

  Let \(p = -5x^2 + 2x - 1\) and \(q = \sqrt{1 - 2x}\).

  Using the quotient rule: \[ \frac{d^2y}{dx^2} = \frac{p'q - pq'}{q^2} \]

  • Find \(p'\): \[ p' = -10x + 2 \]

  • Find \(q'\) (already computed): \[ q' = -\frac{1}{\sqrt{1 - 2x}} \]

  Now substituting: \[ \frac{d^2y}{dx^2} = \frac{(-10x + 2)\sqrt{1 - 2x} - (-5x^2 + 2x - 1)\left(-\frac{1}{\sqrt{1 - 2x}}\right)}{1 - 2x} \]

  Simplifying: \[ = \frac{(-10x + 2)(1 - 2x) + (5x^2 - 2x + 1)}{(1 - 2x)^{3/2}} \]

• Conclusion

Thus, the second derivative is

\[ \frac{d^2y}{dx^2} = \frac{(-10x + 2)(1 - 2x) + (5x^2 - 2x + 1)}{(1 - 2x)^{3/2}} \]

10. We need to find the tenth derivative

\[ \frac{d^{10}y}{dx^{10}} \]

for the function

\[ y = (x^5 - x^4 - x^3 - x^2 - x - 1)(x^5 + 2x^4 + 2x^3 + 2x^2 + 2x + 2). \]

Solution

• Identify the degree of \(y\):

  Each polynomial is of degree 5. The product of two degree 5 polynomials results in a polynomial of degree

  \[ 5 + 5 = 10. \]

• Differentiate \(y\):

  Since \(y\) is a polynomial of degree 10, the tenth derivative will be a constant if the leading term is not zero. The leading term of \(y\) is obtained by multiplying the leading terms of the two polynomials:

  \[ (x^5)(x^5) = x^{10}. \]

• Compute the tenth derivative:

  The \(n\)-th derivative of \(x^n\) is given by:

  \[ \frac{d^n}{dx^n}(x^n) = n!. \]

  Therefore,

  \[ \frac{d^{10}}{dx^{10}}(x^{10}) = 10!. \]

• Conclusion

The tenth derivative is

\[ \frac{d^{10}y}{dx^{10}} = 10!. \]

Thus,

\[ \frac{d^{10}y}{dx^{10}} = 3628800. \]