BSC it Sem 5 Question Papers with Solutions

 here are some important B.Sc. IT Semester 5 questions on successive differentiation along with step-by-step solutions. These problems are designed to help students understand the concepts thoroughly.

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Problem 1: Successive Differentiation of a Polynomial

Problem: Find the first, second, and third derivatives of the function $f(x) = 4x^3 - 3x^2 + 2x - 5$.

Solution:

1. First Derivative:

   • Write the function: $f(x) = 4x^3 - 3x^2 + 2x - 5$
   • Differentiate each term:
     • The derivative of $4x^3$ is $12x^2$
     • The derivative of $-3x^2$ is $-6x$
     • The derivative of $2x$ is $2$
     • The derivative of $-5$ is $0$ (since the derivative of a constant is zero)
   • Combine the results: $$f'(x) = 12x^2 - 6x + 2$$

2. Second Derivative:

   • Write the first derivative: $f'(x) = 12x^2 - 6x + 2$
   • Differentiate each term:
     • The derivative of $12x^2$ is $24x$
     • The derivative of $-6x$ is $-6$
     • The derivative of $2$ is $0$
   • Combine the results: $$f''(x) = 24x - 6$$

3. Third Derivative:

   • Write the second derivative: $f''(x) = 24x - 6$
   • Differentiate each term:
     • The derivative of $24x$ is $24$
     • The derivative of $-6$ is $0$
   • Combine the results: $$f'''(x) = 24$$

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Problem 2: Successive Differentiation of an Exponential Function

Problem: Find the first and second derivatives of the function $g(x) = e^{4x}$.

Solution:

1. First Derivative:

   • Write the function: $g(x) = e^{4x}$
   • Use the chain rule. The outer function is $e^u$ where $u = 4x$:
     • The derivative of $e^u$ is $e^u$
     • Multiply by the derivative of $u$: $\frac{d}{dx}(4x) = 4$
   • Combine the results: $$g'(x) = e^{4x} \cdot 4 = 4e^{4x}$$

2. Second Derivative:

   • Write the first derivative: $g'(x) = 4e^{4x}$
   • Use the chain rule again. The outer function is $e^u$ where $u = 4x$:
     • The derivative of $e^u$ is $e^u$
     • Multiply by the derivative of $u$: $\frac{d}{dx}(4x) = 4$
   • Combine the results: $$g''(x) = 4 \cdot e^{4x} \cdot 4 = 16e^{4x}$$

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Problem 3: Successive Differentiation of a Trigonometric Function

Problem: Find the first, second, and third derivatives of the function $h(x) = \sin(2x)$.

Solution:

1. First Derivative:

   • Write the function: $h(x) = \sin(2x)$
   • Use the chain rule. The outer function is $\sin(u)$ where $u = 2x$:
     • The derivative of $\sin(u)$ is $\cos(u)$
     • Multiply by the derivative of $u$: $\frac{d}{dx}(2x) = 2$
   • Combine the results: $$h'(x) = \cos(2x) \cdot 2 = 2\cos(2x)$$

2. Second Derivative:

   • Write the first derivative: $h'(x) = 2\cos(2x)$
   • Use the chain rule again. The outer function is $\cos(u)$ where $u = 2x$:
     • The derivative of $\cos(u)$ is $-\sin(u)$
     • Multiply by the derivative of $u$: $\frac{d}{dx}(2x) = 2$
   • Combine the results: $$h''(x) = 2 \cdot -\sin(2x) \cdot 2 = -4\sin(2x)$$

3. Third Derivative:

   • Write the second derivative: $h''(x) = -4\sin(2x)$
   • Use the chain rule again. The outer function is $\sin(u)$ where $u = 2x$:
     • The derivative of $\sin(u)$ is $\cos(u)$
     • Multiply by the derivative of $u$: $\frac{d}{dx}(2x) = 2$
   • Combine the results: $$h'''(x) = -4 \cdot \cos(2x) \cdot 2 = -8\cos(2x)$$

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Problem 4: Successive Differentiation of a Logarithmic Function

Problem: Find the first and second derivatives of the function $k(x) = \ln(2x)$.

Solution:

1. First Derivative:

   • Write the function: $k(x) = \ln(2x)$
   • Use the chain rule. The outer function is $\ln(u)$ where $u = 2x$:
     • The derivative of $\ln(u)$ is $\frac{1}{u}$
     • Multiply by the derivative of $u$: $\frac{d}{dx}(2x) = 2$
   • Combine the results: $$k'(x) = \frac{1}{2x} \cdot 2 = \frac{2}{2x} = \frac{1}{x}$$

2. Second Derivative:

   • Write the first derivative: $k'(x) = \frac{1}{x}$
   • Rewrite $\frac{1}{x}$ as $x^{-1}$
   • The derivative of $x^{-1}$ is $-x^{-2}$
   • Combine the results: $$k''(x) = -\frac{1}{x^2}$$

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Problem 5: Successive Differentiation of a Rational Function

Problem: Find the first and second derivatives of the function $m(x) = \frac{1}{x+2}$.

Solution:

1. First Derivative:

   • Write the function: $m(x) = \frac{1}{x+2}$
   • Rewrite $\frac{1}{x+2}$ as $(x+2)^{-1}$
   • Use the power rule and chain rule:
     • The derivative of $(x+2)^{-1}$ is $-1 \cdot (x+2)^{-2}$
     • Multiply by the derivative of $x+2$, which is 1
   • Combine the results: $$m'(x) = -\frac{1}{(x+2)^2}$$

2. Second Derivative:

   • Write the first derivative: $m'(x) = -\frac{1}{(x+2)^2}$
   • Rewrite $-\frac{1}{(x+2)^2}$ as $-(x+2)^{-2}$
   • Use the power rule and chain rule again:
     • The derivative of $-(x+2)^{-2}$ is $2(x+2)^{-3}$
     • Multiply by the derivative of $x+2$, which is 1
   • Combine the results: $$m''(x) = \frac{2}{(x+2)^3}$$

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Problem 6: Successive Differentiation of a Combined Function

Problem: Find the first and second derivatives of the function $n(x) = x^2 \cdot \ln(x)$.

Solution:

1. First Derivative:

   • Write the function: $n(x) = x^2 \cdot \ln(x)$
   • Use the product rule: $(uv)' = u'v + uv'$
     • Let $u = x^2$ and $v = \ln(x)$
     • The derivative of $u = x^2$ is $u' = 2x$
     • The derivative of $v = \ln(x)$ is $v' = \frac{1}{x}$
   • Combine the results: $$n'(x) = (2x) \cdot \ln(x) + (x^2) \cdot \frac{1}{x} = 2x \ln(x) + x$$

2. Second Derivative:

• Write the first derivative: $n'(x) = 2x \ln(x) + x$
• Differentiate each term using the product rule again for $2x \ln(x)$:
  • Let $u = 2x$ and $v = \ln(x)$
  • The derivative of $u = 2x$ is $u' = 2$
  • The derivative of $v = \ln(x)$ is $v' = \frac{1}{x}$
  • Combine the results for the first term: $$(2x \ln(x))' = (2) \cdot \ln(x) + (2x) \cdot \frac{1}{x} = 2 \ln(x) + 2$$
  • The derivative of the second term $x$ is $1$
• Combine all results: $$n''(x) = 2 \ln(x) + 2 + 1 = 2 \ln(x) + 3$$

Problem 7: Successive Differentiation of a Trigonometric Function

Problem: Find the first, second, and third derivatives of the function $f(x) = \cos(3x)$.

Solution:

1. First Derivative:

   • Write the function: $f(x) = \cos(3x)$
   • Use the chain rule. The outer function is $\cos(u)$ where $u = 3x$:
     • The derivative of $\cos(u)$ is $-\sin(u)$
     • Multiply by the derivative of $u$: $\frac{d}{dx}(3x) = 3$
   • Combine the results: $$f'(x) = -\sin(3x) \cdot 3 = -3\sin(3x)$$

2. Second Derivative:

   • Write the first derivative: $f'(x) = -3\sin(3x)$
   • Use the chain rule again. The outer function is $\sin(u)$ where $u = 3x$:
     • The derivative of $\sin(u)$ is $\cos(u)$
     • Multiply by the derivative of $u$: $\frac{d}{dx}(3x) = 3$
   • Combine the results: $$f''(x) = -3 \cdot \cos(3x) \cdot 3 = -9\cos(3x)$$

3. Third Derivative:

   • Write the second derivative: $f''(x) = -9\cos(3x)$
   • Use the chain rule again. The outer function is $\cos(u)$ where $u = 3x$:
     • The derivative of $\cos(u)$ is $-\sin(u)$
     • Multiply by the derivative of $u$: $\frac{d}{dx}(3x) = 3$
   • Combine the results: $$f'''(x) = -9 \cdot -\sin(3x) \cdot 3 = 27\sin(3x)$$

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Problem 8: Successive Differentiation of a Logarithmic Function

Problem: Find the first and second derivatives of the function $g(x) = \ln(5x)$.

Solution:

1. First Derivative:

   • Write the function: $g(x) = \ln(5x)$
   • Use the chain rule. The outer function is $\ln(u)$ where $u = 5x$:
     • The derivative of $\ln(u)$ is $\frac{1}{u}$
     • Multiply by the derivative of $u$: $\frac{d}{dx}(5x) = 5$
   • Combine the results: $$g'(x) = \frac{1}{5x} \cdot 5 = \frac{5}{5x} = \frac{1}{x}$$

2. Second Derivative:

   • Write the first derivative: $g'(x) = \frac{1}{x}$
   • Rewrite $\frac{1}{x}$ as $x^{-1}$
   • The derivative of $x^{-1}$ is $-x^{-2}$
   • Combine the results: $$g''(x) = -\frac{1}{x^2}$$

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Problem 9: Successive Differentiation of a Polynomial Function

Problem: Find the first, second, and third derivatives of the function $h(x) = 2x^4 - x^3 + 3x - 7$.

Solution:

1. First Derivative:

   • Write the function: $h(x) = 2x^4 - x^3 + 3x - 7$
   • Differentiate each term:
     • The derivative of $2x^4$ is $8x^3$
     • The derivative of $-x^3$ is $-3x^2$
     • The derivative of $3x$ is $3$
     • The derivative of $-7$ is $0$
   • Combine the results: $$h'(x) = 8x^3 - 3x^2 + 3$$

2. Second Derivative:

   • Write the first derivative: $h'(x) = 8x^3 - 3x^2 + 3$
   • Differentiate each term:
     • The derivative of $8x^3$ is $24x^2$
     • The derivative of $-3x^2$ is $-6x$
     • The derivative of $3$ is $0$
   • Combine the results: $$h''(x) = 24x^2 - 6x$$

3. Third Derivative:

   • Write the second derivative: $h''(x) = 24x^2 - 6x$
   • Differentiate each term:
     • The derivative of $24x^2$ is $48x$
     • The derivative of $-6x$ is $-6$
   • Combine the results: $$h'''(x) = 48x - 6$$

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Problem 10: Successive Differentiation of an Exponential Function

Problem: Find the first and second derivatives of the function $k(x) = e^{2x}$.

Solution:

1. First Derivative:

   • Write the function: $k(x) = e^{2x}$
   • Use the chain rule. The outer function is $e^u$ where $u = 2x$:
     • The derivative of $e^u$ is $e^u$
     • Multiply by the derivative of $u$: $\frac{d}{dx}(2x) = 2$
   • Combine the results: $$k'(x) = e^{2x} \cdot 2 = 2e^{2x}$$

2. Second Derivative:

   • Write the first derivative: $k'(x) = 2e^{2x}$
   • Use the chain rule again. The outer function is $e^u$ where $u = 2x$:
     • The derivative of $e^u$ is $e^u$
     • Multiply by the derivative of $u$: $\frac{d}{dx}(2x) = 2$
   • Combine the results: $$k''(x) = 2 \cdot e^{2x} \cdot 2 = 4e^{2x}$$

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Problem 11: Successive Differentiation of a Rational Function

Problem: Find the first and second derivatives of the function $m(x) = \frac{1}{2x+3}$.

Solution:

1. First Derivative:

   • Write the function: $m(x) = \frac{1}{2x+3}$
   • Rewrite $\frac{1}{2x+3}$ as $(2x+3)^{-1}$
   • Use the power rule and chain rule:
     • The derivative of $(2x+3)^{-1}$ is $-1 \cdot (2x+3)^{-2}$
     • Multiply by the derivative of $2x+3$, which is 2
   • Combine the results: $$m'(x) = -\frac{1}{(2x+3)^2} \cdot 2 = -\frac{2}{(2x+3)^2}$$

2. Second Derivative:

   • Write the first derivative: $m'(x) = -\frac{2}{(2x+3)^2}$
   • Rewrite $-\frac{2}{(2x+3)^2}$ as $-2(2x+3)^{-2}$
   • Use the power rule and chain rule again:
     • The derivative of $-2(2x+3)^{-2}$ is $-2 \cdot -2 \cdot (2x+3)^{-3}$
     • Multiply by the derivative of $2x+3$, which is 2
   • Combine the results: $$m''(x) = 4 \cdot \frac{1}{(2x+3)^3} = \frac{8}{(2x+3)^3}$$

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Problem 12: Successive Differentiation of a Combined Function

Problem: Find the first and second derivatives of the function $n(x) = x^3 \cdot e^x$.

Solution:

1. First Derivative:

   • Write the function: $n(x) = x^3 \cdot e^x$
   • Use the product rule: $(uv)' = u'v + uv'$
     • Let $u = x^3$ and $v = e^x$
     • The derivative of $u = x^3$ is $u' = 3x^2$
     • The derivative of $v = e^x$ is $v' = e^x$
   • Combine the results: $$n'(x) = (3x^2) \cdot e^x + (x^3) \cdot e^x = 3x^2 e^x + x^3 e^x$$
   • Factor out the common term $e^x$: $$n'(x) = e^x (3x^2 + x^3) = e^x x^2 (3 + x)$$

2. Second Derivative:

   • Write the first derivative: $n'(x) = e^x (3x^2 + x^3)$
   • Use the product rule again:
     • Let $u = e^x$ and $v = 3x^2 + x^3$
     • The derivative of $u = e^x$ is $u' = e^x$
     • The derivative of $v = 3x^2 + x^3$ is $v' = 6x + 3x^2$
   • Combine the results: $$n''(x) = (e^x) \cdot (6x + 3x^2) + (e^x) \cdot (3x^2 + x^3) = e^x (6x + 3x^2 + 3x^2 + x^3) = e^x (6x + 6x^2 + x^3)$$
   • Simplify the expression: $$n''(x) = e^x (x^3 + 6x^2 + 6x)$$