Rbse Solutions Class 12 Maths Chapter 5 Exercise 5.7: Second Order Derivatives

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Full solutions for Class 12 Maths (NCERT) Exercise 5.7. Learn to find the second order derivative ($\frac{d^2 y}{dx^2}$) for various functions using the Chain, Product, and Quotient rules, and prove identities involving derivatives like $\frac{d^2 y}{dx^2} + y = 0$.

Rbse Solutions Class 12 Maths Chapter 5 Exercise 5.7: Second Order Derivatives

The second order derivative of a function $y = f(x)$ is the derivative of the first derivative, denoted as $\frac{d^2 y}{dx^2}$ or $y”$.

Finding Second Order Derivatives (Exercises 1 – 10)

1. $y = x^2 + 3x + 2$

  1. First Derivative ($\frac{dy}{dx}$ or $y’$):$$y’ = 2x + 3$$
  2. Second Derivative ($\frac{d^2 y}{dx^2}$ or $y”$):$$y” = \frac{d}{dx}(2x + 3) = 2$$$$\mathbf{\frac{d^2 y}{dx^2} = 2}$$

2. $y = x^{20}$

  1. First Derivative:$$y’ = 20x^{19}$$
  2. Second Derivative:$$y” = \frac{d}{dx}(20x^{19}) = 20 \cdot 19x^{18} = 380x^{18}$$$$\mathbf{\frac{d^2 y}{dx^2} = 380x^{18}}$$

3. $y = x \cos x$

Use the Product Rule: $\frac{d}{dx}(uv) = u’v + uv’$.

  1. First Derivative:$$y’ = \frac{d}{dx}(x) \cdot \cos x + x \cdot \frac{d}{dx}(\cos x)$$$$y’ = 1 \cdot \cos x + x (-\sin x) = \cos x – x \sin x$$
  2. Second Derivative: Differentiate $y’$ (use the Product Rule again for $-x \sin x$).$$y” = \frac{d}{dx}(\cos x) – \left[ \frac{d}{dx}(x) \sin x + x \frac{d}{dx}(\sin x) \right]$$$$y” = -\sin x – \left[ 1 \cdot \sin x + x \cos x \right]$$$$y” = -\sin x – \sin x – x \cos x$$$$\mathbf{\frac{d^2 y}{dx^2} = -2 \sin x – x \cos x}$$

4. $y = \log x$

  1. First Derivative:$$y’ = \frac{1}{x} = x^{-1}$$
  2. Second Derivative:$$y” = \frac{d}{dx}(x^{-1}) = -1 x^{-2} = -\frac{1}{x^2}$$$$\mathbf{\frac{d^2 y}{dx^2} = -\frac{1}{x^2}}$$

5. $y = x^3 \log x$

Use the Product Rule.

  1. First Derivative:$$y’ = \frac{d}{dx}(x^3) \cdot \log x + x^3 \cdot \frac{d}{dx}(\log x)$$$$y’ = 3x^2 \log x + x^3 \cdot \frac{1}{x} = 3x^2 \log x + x^2$$
  2. Second Derivative: Differentiate $y’$ (use the Product Rule again for $3x^2 \log x$).$$y” = \frac{d}{dx}(3x^2 \log x) + \frac{d}{dx}(x^2)$$$$y” = \left[ 6x \log x + 3x^2 \cdot \frac{1}{x} \right] + 2x$$$$y” = 6x \log x + 3x + 2x = 6x \log x + 5x$$$$\mathbf{\frac{d^2 y}{dx^2} = x (6 \log x + 5)}$$

6. $y = e^x \sin 5x$

Use the Product Rule.

  1. First Derivative:$$y’ = \frac{d}{dx}(e^x) \sin 5x + e^x \frac{d}{dx}(\sin 5x)$$$$y’ = e^x \sin 5x + e^x (5 \cos 5x) = e^x (\sin 5x + 5 \cos 5x)$$
  2. Second Derivative: Use the Product Rule on $y’$. Let $u = e^x$ and $v = (\sin 5x + 5 \cos 5x)$.$$y” = \frac{d}{dx}(e^x) v + e^x \frac{d}{dx}(v)$$$$\frac{d}{dx}(v) = \frac{d}{dx}(\sin 5x + 5 \cos 5x) = 5 \cos 5x – 5(5 \sin 5x) = 5 \cos 5x – 25 \sin 5x$$$$y” = e^x (\sin 5x + 5 \cos 5x) + e^x (5 \cos 5x – 25 \sin 5x)$$$$y” = e^x [\sin 5x – 25 \sin 5x + 5 \cos 5x + 5 \cos 5x]$$$$y” = e^x [-24 \sin 5x + 10 \cos 5x]$$$$\mathbf{\frac{d^2 y}{dx^2} = 2e^x (5 \cos 5x – 12 \sin 5x)}$$

7. $y = e^{6x} \cos 3x$

Use the Product Rule.

  1. First Derivative:$$y’ = \frac{d}{dx}(e^{6x}) \cos 3x + e^{6x} \frac{d}{dx}(\cos 3x)$$$$y’ = (6e^{6x}) \cos 3x + e^{6x} (-3 \sin 3x) = e^{6x} (6 \cos 3x – 3 \sin 3x)$$
  2. Second Derivative: Use the Product Rule on $y’$. Let $u = e^{6x}$ and $v = (6 \cos 3x – 3 \sin 3x)$.$$y” = \frac{d}{dx}(e^{6x}) v + e^{6x} \frac{d}{dx}(v)$$$$\frac{d}{dx}(v) = -6(3 \sin 3x) – 3(3 \cos 3x) = -18 \sin 3x – 9 \cos 3x$$$$y” = (6e^{6x}) (6 \cos 3x – 3 \sin 3x) + e^{6x} (-18 \sin 3x – 9 \cos 3x)$$$$y” = e^{6x} [36 \cos 3x – 18 \sin 3x – 18 \sin 3x – 9 \cos 3x]$$$$y” = e^{6x} [27 \cos 3x – 36 \sin 3x]$$$$\mathbf{\frac{d^2 y}{dx^2} = 9e^{6x} (3 \cos 3x – 4 \sin 3x)}$$

8. $y = \tan^{-1} x$

  1. First Derivative:$$y’ = \frac{1}{1 + x^2} = (1 + x^2)^{-1}$$
  2. Second Derivative: Use the Chain Rule on $y’$.$$y” = \frac{d}{dx}((1 + x^2)^{-1}) = -1 (1 + x^2)^{-2} \cdot \frac{d}{dx}(1 + x^2)$$$$y” = – (1 + x^2)^{-2} \cdot (2x) = -\frac{2x}{(1 + x^2)^2}$$$$\mathbf{\frac{d^2 y}{dx^2} = -\frac{2x}{(1 + x^2)^2}}$$

9. $y = \log (\log x)$

Use the Chain Rule.

  1. First Derivative:$$y’ = \frac{1}{\log x} \cdot \frac{d}{dx}(\log x) = \frac{1}{\log x} \cdot \frac{1}{x} = (x \log x)^{-1}$$
  2. Second Derivative: Use the Chain Rule or Quotient Rule on $y’ = \frac{1}{x \log x}$.Using the Quotient Rule (let $u=1, v=x \log x$):$$y” = \frac{v u’ – u v’}{v^2} = \frac{(x \log x) \cdot 0 – 1 \cdot \frac{d}{dx}(x \log x)}{(x \log x)^2}$$$$y” = – \frac{1 \cdot \log x + x \cdot \frac{1}{x}}{(x \log x)^2} = – \frac{\log x + 1}{(x \log x)^2}$$$$\mathbf{\frac{d^2 y}{dx^2} = -\frac{1 + \log x}{(x \log x)^2}}$$

10. $y = \sin (\log x)$

Use the Chain Rule.

  1. First Derivative:$$y’ = \cos (\log x) \cdot \frac{d}{dx}(\log x) = \frac{\cos (\log x)}{x}$$
  2. Second Derivative: Use the Quotient Rule on $y’$.$$y” = \frac{x \cdot \frac{d}{dx}(\cos (\log x)) – \cos (\log x) \cdot \frac{d}{dx}(x)}{x^2}$$$$y” = \frac{x \cdot \left( -\sin (\log x) \cdot \frac{1}{x} \right) – \cos (\log x) \cdot 1}{x^2}$$$$y” = \frac{-\sin (\log x) – \cos (\log x)}{x^2}$$$$\mathbf{\frac{d^2 y}{dx^2} = -\frac{\sin (\log x) + \cos (\log x)}{x^2}}$$

Proof Questions (Exercises 11 – 17)

11. If $y = 5 \cos x – 3 \sin x$, prove that $\frac{d^2 y}{dx^2} + y = 0$.

  1. First Derivative:$$\frac{dy}{dx} = 5 (-\sin x) – 3 (\cos x) = -5 \sin x – 3 \cos x$$
  2. Second Derivative:$$\frac{d^2 y}{dx^2} = -5 (\cos x) – 3 (-\sin x) = -5 \cos x + 3 \sin x$$
  3. Substitute into the equation $\frac{d^2 y}{dx^2} + y = 0$:$$(-5 \cos x + 3 \sin x) + (5 \cos x – 3 \sin x) = 0$$$$0 = 0$$Hence Proved.

12. If $y = \cos^{-1} x$, Find $\frac{d^2 y}{dx^2}$ in terms of $y$ alone.

  1. First Derivative:$$\frac{dy}{dx} = -\frac{1}{\sqrt{1 – x^2}}$$
  2. Relate $x$ to $y$: Since $y = \cos^{-1} x$, we have $x = \cos y$.$$\sqrt{1 – x^2} = \sqrt{1 – \cos^2 y} = \sqrt{\sin^2 y} = \sin y$$Substituting back into $\frac{dy}{dx}$:$$\frac{dy}{dx} = -\frac{1}{\sin y} = -\csc y$$
  3. Second Derivative (in terms of $y$): Differentiate $\frac{dy}{dx}$ w.r.t. $x$.$$\frac{d^2 y}{dx^2} = \frac{d}{dx} (-\csc y) = \frac{d}{dy} (-\csc y) \cdot \frac{dy}{dx}$$$$\frac{d^2 y}{dx^2} = – (-\csc y \cot y) \cdot \frac{dy}{dx}$$$$\frac{d^2 y}{dx^2} = \csc y \cot y \cdot (-\csc y)$$ (Substituting $\frac{dy}{dx} = -\csc y$)$$\mathbf{\frac{d^2 y}{dx^2} = -\csc^2 y \cot y}$$

13. If $y = 3 \cos (\log x) + 4 \sin (\log x)$, show that $x^2 y_2 + xy_1 + y = 0$.

($y_1 = \frac{dy}{dx}$ and $y_2 = \frac{d^2 y}{dx^2}$)

  1. First Derivative ($y_1$):$$y_1 = 3 (-\sin (\log x)) \cdot \frac{1}{x} + 4 (\cos (\log x)) \cdot \frac{1}{x}$$$$y_1 = \frac{1}{x} [-3 \sin (\log x) + 4 \cos (\log x)]$$Multiply by $x$:$$xy_1 = -3 \sin (\log x) + 4 \cos (\log x)$$ (Equation A)
  2. Second Derivative ($y_2$): Differentiate Equation A w.r.t. $x$ (using the Product Rule on $xy_1$):$$\frac{d}{dx}(x y_1) = \frac{d}{dx}(-3 \sin (\log x) + 4 \cos (\log x))$$$$1 \cdot y_1 + x \cdot y_2 = -3 (\cos (\log x)) \cdot \frac{1}{x} + 4 (-\sin (\log x)) \cdot \frac{1}{x}$$$$y_1 + x y_2 = \frac{1}{x} [-3 \cos (\log x) – 4 \sin (\log x)]$$Multiply by $x$:$$xy_1 + x^2 y_2 = – [3 \cos (\log x) + 4 \sin (\log x)]$$Recognize the term in the brackets as the original function $y$:$$xy_1 + x^2 y_2 = -y$$Rearrange:$$\mathbf{x^2 y_2 + xy_1 + y = 0}$$Hence Proved.

14. If $y = Ae^{mx} + Be^{nx}$, show that $\frac{d^2 y}{dx^2} – (m + n) \frac{dy}{dx} + mny = 0$.

  1. First Derivative ($\frac{dy}{dx}$):$$\frac{dy}{dx} = A (me^{mx}) + B (ne^{nx}) = Ame^{mx} + Bne^{nx}$$
  2. Second Derivative ($\frac{d^2 y}{dx^2}$):$$\frac{d^2 y}{dx^2} = Am (me^{mx}) + Bn (ne^{nx}) = Am^2 e^{mx} + Bn^2 e^{nx}$$
  3. Substitute into the expression $\frac{d^2 y}{dx^2} – (m + n) \frac{dy}{dx} + mny$:$$[Am^2 e^{mx} + Bn^2 e^{nx}] – (m + n) [Ame^{mx} + Bne^{nx}] + mn [Ae^{mx} + Be^{nx}]$$Expand the middle term:$$= Am^2 e^{mx} + Bn^2 e^{nx} – (m^2 Ae^{mx} + mne^{nx} + nmAe^{mx} + n^2 Be^{nx}) + mnAe^{mx} + mnBe^{nx}$$$$= Am^2 e^{mx} + Bn^2 e^{nx} – Am^2 e^{mx} – mnAe^{mx} – mnBe^{nx} – n^2 Be^{nx} + mnAe^{mx} + mnBe^{nx}$$Group terms (all terms cancel out):$$(Am^2 e^{mx} – Am^2 e^{mx}) + (Bn^2 e^{nx} – Bn^2 e^{nx}) + (-mnAe^{mx} + mnAe^{mx}) + (-mnBe^{nx} + mnBe^{nx}) = 0$$Hence Proved.

15. If $y = 500e^{7x} + 600e^{-7x}$, show that $\frac{d^2 y}{dx^2} = 49y$.

  1. First Derivative:$$\frac{dy}{dx} = 500 (7e^{7x}) + 600 (-7e^{-7x}) = 3500e^{7x} – 4200e^{-7x}$$
  2. Second Derivative:$$\frac{d^2 y}{dx^2} = 3500 (7e^{7x}) – 4200 (-7e^{-7x})$$$$\frac{d^2 y}{dx^2} = 24500e^{7x} + 29400e^{-7x}$$
  3. Factor out 49:$$\frac{d^2 y}{dx^2} = 49 (500e^{7x} + 600e^{-7x})$$Recognize the term in the brackets as the original function $y$:$$\mathbf{\frac{d^2 y}{dx^2} = 49y}$$Hence Proved.

16. If $e^y (x + 1) = 1$, show that $\frac{d^2 y}{dx^2} = \left(\frac{dy}{dx}\right)^2$.

  1. Isolate $e^y$:$$e^y = \frac{1}{x + 1} = (x + 1)^{-1}$$
  2. First Derivative ($\frac{dy}{dx}$): Differentiate implicitly w.r.t. $x$.$$\frac{d}{dx}(e^y (x + 1)) = \frac{d}{dx}(1)$$ (Using Product Rule on LHS)$$e^y \frac{dy}{dx} (x + 1) + e^y (1) = 0$$Divide by $e^y$:$$\frac{dy}{dx} (x + 1) + 1 = 0$$$$\frac{dy}{dx} = -\frac{1}{x + 1} = -(x + 1)^{-1}$$
  3. Second Derivative ($\frac{d^2 y}{dx^2}$): Differentiate $\frac{dy}{dx}$ w.r.t. $x$.$$\frac{d^2 y}{dx^2} = \frac{d}{dx} (-(x + 1)^{-1})$$$$\frac{d^2 y}{dx^2} = – (-1) (x + 1)^{-2} \cdot 1 = (x + 1)^{-2}$$$$\frac{d^2 y}{dx^2} = \frac{1}{(x + 1)^2}$$
  4. Compare $\frac{d^2 y}{dx^2}$ with $\left(\frac{dy}{dx}\right)^2$:$$\left(\frac{dy}{dx}\right)^2 = \left( -\frac{1}{x + 1} \right)^2 = \frac{1}{(x + 1)^2}$$Since $\frac{d^2 y}{dx^2} = \frac{1}{(x + 1)^2}$ and $\left(\frac{dy}{dx}\right)^2 = \frac{1}{(x + 1)^2}$, we have:$$\mathbf{\frac{d^2 y}{dx^2} = \left(\frac{dy}{dx}\right)^2}$$Hence Proved.

17. If $y = (\tan^{-1} x)^2$, show that $(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2$.

  1. First Derivative ($y_1$):$$y_1 = 2 (\tan^{-1} x) \cdot \frac{1}{1 + x^2}$$Multiply by $(1 + x^2)$:$$(1 + x^2) y_1 = 2 \tan^{-1} x$$ (Equation B)
  2. Second Derivative ($y_2$): Differentiate Equation B w.r.t. $x$ (using the Product Rule on LHS):$$\frac{d}{dx}((1 + x^2) y_1) = \frac{d}{dx}(2 \tan^{-1} x)$$$$(2x) y_1 + (1 + x^2) y_2 = 2 \cdot \frac{1}{1 + x^2}$$Multiply the entire equation by $(1 + x^2)$:$$(1 + x^2) [2x y_1 + (1 + x^2) y_2] = 2$$$$2x (1 + x^2) y_1 + (1 + x^2)^2 y_2 = 2$$Rearrange to match the required form:$$\mathbf{(x^2 + 1)^2 y_2 + 2x (x^2 + 1) y_1 = 2}$$Hence Proved.
Ex 5.1
Ex 5.2
Ex 5.3
Ex 5.4
Ex 5.5
Ex 5.6
Ex 5.7
Ch 5 Misc.

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