Exercise 5.4, Question 1
\(\dfrac{e^x}{\sin x}\)
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- Quotient rule: \(\dfrac{e^x\sin x - e^x\cos x}{\sin^2 x}\).
Answer: \(\dfrac{e^x(\sin x - \cos x)}{\sin^2 x}\), \(x \ne n\pi\)
NCERT Solutions for Class 12 Maths Chapter 5 Exercise 5.4
Exercise 5.4: Exponential and logarithmic functions. \((e^x)' = e^x\), \((\log x)' = \tfrac1x\) (x > 0), combined with the chain, product and quotient rules: \((e^{u})' = e^{u}u'\), \((\log u)' = \tfrac{u'}{u}\).
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Exercise 5.4 questions and solutions
Exercise 5.4, Question 2
\(e^{\sin^{-1}x}\)
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- \[e^{\sin^{-1}x} \cdot \dfrac{1}{\sqrt{1 - x^2}}\]
Answer: \(\dfrac{e^{\sin^{-1}x}}{\sqrt{1 - x^2}}\), \(-1 < x < 1\)
Exercise 5.4, Question 3
\(e^{x^3}\)
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- \(e^{x^3} \cdot 3x^2\).
Answer: \(3x^2e^{x^3}\)
Exercise 5.4, Question 4
\(\sin(\tan^{-1}e^{-x})\)
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- \[\cos(\tan^{-1}e^{-x}) \cdot \dfrac{1}{1 + e^{-2x}} \cdot (-e^{-x})\]
Answer: \[-\dfrac{e^{-x}\cos(\tan^{-1}e^{-x})}{1 + e^{-2x}}\]
Exercise 5.4, Question 5
\(\log(\cos e^x)\)
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- \[\dfrac{1}{\cos e^x} \cdot (-\sin e^x) \cdot e^x\]
Answer: \(-e^x\tan e^x\), \(e^x \ne (2n + 1)\tfrac{\pi}{2}\)
Exercise 5.4, Question 6
\(e^x + e^{x^2} + \dots + e^{x^5}\)
Show solution
- Term by term: \((e^{x^k})' = kx^{k-1}e^{x^k}\).
Answer: \[e^x + 2xe^{x^2} + 3x^2e^{x^3} + 4x^3e^{x^4} + 5x^4e^{x^5}\]
Exercise 5.4, Question 7
\(\sqrt{e^{\sqrt x}}\), \(x > 0\)
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- \(\sqrt{e^{\sqrt x}} = e^{\sqrt x/2}\), so the derivative is \(e^{\sqrt x/2} \cdot \dfrac{1}{4\sqrt x}\).
Answer: \(\dfrac{\sqrt{e^{\sqrt x}}}{4\sqrt x}\)
Exercise 5.4, Question 8
\(\log(\log x)\), \(x > 1\)
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- \(\dfrac{1}{\log x} \cdot \dfrac1x\).
Answer: \(\dfrac{1}{x\log x}\)
Exercise 5.4, Question 9
\(\dfrac{\cos x}{\log x}\), \(x > 0\)
Show solution
- Quotient rule: \[\dfrac{-\sin x\log x - \cos x \cdot \frac1x}{(\log x)^2}\]
Answer: \[-\dfrac{x\log x\sin x + \cos x}{x(\log x)^2}\]
Exercise 5.4, Question 10
\(\cos(\log x + e^x)\), \(x > 0\)
Show solution
- \[-\sin(\log x + e^x) \cdot \left(\tfrac1x + e^x\right)\]
Answer: \[-\left(\dfrac1x + e^x\right)\sin(\log x + e^x)\]
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