Exercise 5.7, Question 1
\(x^2 + 3x + 2\)
Show solution
- \(y' = 2x + 3\).
Answer: \(y'' = 2\)
NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.7
Exercise 5.7: Second order derivatives. Differentiate twice: \(y'' = \tfrac{d}{dx}(y')\). To prove a relation between \(y, y', y''\), it is often quicker to clear roots or fractions after the first derivative (e.g. \(\sqrt{1 - x^2}\,y' = 1\)) and differentiate that.
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Exercise 5.7 questions and solutions
Exercise 5.7, Question 2
\(x^{20}\)
Show solution
- \(y' = 20x^{19}\).
Answer: \(y'' = 380x^{18}\)
Exercise 5.7, Question 3
\(x\cos x\)
Show solution
- \(y' = \cos x - x\sin x\).
Answer: \(y'' = -2\sin x - x\cos x\)
Exercise 5.7, Question 4
\(\log x\)
Show solution
- \(y' = \dfrac1x\).
Answer: \(y'' = -\dfrac{1}{x^2}\)
Exercise 5.7, Question 5
\(x^3\log x\)
Show solution
- \(y' = 3x^2\log x + x^2\).
- \(y'' = 6x\log x + 3x + 2x\).
Answer: \(y'' = x(5 + 6\log x)\)
Exercise 5.7, Question 6
\(e^x\sin 5x\)
Show solution
- \(y' = e^x(\sin 5x + 5\cos 5x)\).
- \[y'' = e^x(\sin 5x + 5\cos 5x) + e^x(5\cos 5x - 25\sin 5x)\]
Answer: \(y'' = 2e^x(5\cos 5x - 12\sin 5x)\)
Exercise 5.7, Question 7
\(e^{6x}\cos 3x\)
Show solution
- \(y' = e^{6x}(6\cos 3x - 3\sin 3x)\).
- \[y'' = 6e^{6x}(6\cos 3x - 3\sin 3x) + e^{6x}(-18\sin 3x - 9\cos 3x)\]
Answer: \(y'' = 9e^{6x}(3\cos 3x - 4\sin 3x)\)
Exercise 5.7, Question 8
\(\tan^{-1}x\)
Show solution
- \(y' = (1 + x^2)^{-1}\).
Answer: \(y'' = -\dfrac{2x}{(1 + x^2)^2}\)
Exercise 5.7, Question 9
\(\log(\log x)\)
Show solution
- \(y' = (x\log x)^{-1}\).
- \(y'' = -(x\log x)^{-2}(\log x + 1)\).
Answer: \(y'' = -\dfrac{1 + \log x}{(x\log x)^2}\)
Exercise 5.7, Question 10
\(\sin(\log x)\)
Show solution
- \(y' = \dfrac{\cos(\log x)}{x}\).
- \[y'' = \dfrac{-\sin(\log x) \cdot \frac1x \cdot x - \cos(\log x)}{x^2}\]
Answer: \[y'' = -\dfrac{\sin(\log x) + \cos(\log x)}{x^2}\]
Exercise 5.7, Question 11
\(y = 5\cos x - 3\sin x\). Prove \(y'' + y = 0\).
Show solution
- \(y' = -5\sin x - 3\cos x\), \(y'' = -5\cos x + 3\sin x = -y\).
Answer: Proved
Exercise 5.7, Question 12
\(y = \cos^{-1}x\). Find \(y''\) in terms of y alone.
Show solution
- \(y' = -(1 - x^2)^{-1/2}\), \(y'' = -\dfrac{x}{(1 - x^2)^{3/2}}\).
- \(x = \cos y\) and \(\sqrt{1 - x^2} = \sin y\) (as \(y \in [0, \pi]\)): \(y'' = -\dfrac{\cos y}{\sin^3 y}\).
Answer: \(y'' = -\cot y\csc^2 y\)
Exercise 5.7, Question 13
\(y = 3\cos(\log x) + 4\sin(\log x)\). Show \(x^2y'' + xy' + y = 0\).
Show solution
- \[y' = \dfrac{-3\sin(\log x) + 4\cos(\log x)}{x}\], so \(xy' = -3\sin(\log x) + 4\cos(\log x)\).
- Differentiate \(xy'\): \[\begin{aligned}xy'' + y' &= \dfrac{-3\cos(\log x) - 4\sin(\log x)}{x} \\ &= -\dfrac{y}{x}\end{aligned}\]
- Multiply by x: \(x^2y'' + xy' + y = 0\).
Answer: Shown
Exercise 5.7, Question 14
\(y = Ae^{mx} + Be^{nx}\). Show \(y'' - (m + n)y' + mny = 0\).
Show solution
- \(y' = Ame^{mx} + Bne^{nx}\), \(y'' = Am^2e^{mx} + Bn^2e^{nx}\).
- Coefficient of \(Ae^{mx}\): \(m^2 - (m + n)m + mn = 0\); of \(Be^{nx}\): \(n^2 - (m + n)n + mn = 0\).
Answer: Shown
Exercise 5.7, Question 15
\(y = 500e^{7x} + 600e^{-7x}\). Show \(y'' = 49y\).
Show solution
- \(y' = 3500e^{7x} - 4200e^{-7x}\), \[\begin{aligned}y'' &= 24500e^{7x} + 29400e^{-7x} \\ &= 49(500e^{7x} + 600e^{-7x})\end{aligned}\]
Answer: Shown
Exercise 5.7, Question 16
\(e^y(x + 1) = 1\). Show \(y'' = (y')^2\).
Show solution
- \(e^y = \dfrac{1}{x + 1}\), so \(y = -\log(x + 1)\).
- \(y' = -\dfrac{1}{x + 1}\), \(y'' = \dfrac{1}{(x + 1)^2} = (y')^2\).
Answer: Shown
Exercise 5.7, Question 17
\(y = (\tan^{-1}x)^2\). Show \((x^2 + 1)^2y'' + 2x(x^2 + 1)y' = 2\).
Show solution
- \(y' = \dfrac{2\tan^{-1}x}{1 + x^2}\), so \((1 + x^2)y' = 2\tan^{-1}x\).
- Differentiate: \((1 + x^2)y'' + 2xy' = \dfrac{2}{1 + x^2}\).
- Multiply by \(1 + x^2\).
Answer: Shown
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