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CONTINUITY AND DIFFERENTIABILITY — Maths STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceMathsCONTINUITY AND DIFFERENTIABILITY4 Marks
Question
If y = f(u) is a differentiable function of u and u = g(x) is a differentiable function of x, then y = f(g(x)] is a differentiable function of x and $\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{du}}\times\frac{\text{du}}{\text{dx}}.$ This rule is also known as CHAIN RULE.
Based on the above information, find the derivative of functions w.r.t. x in the following questions.
  1. $\cos\sqrt{\text{x}}$
  1. $\frac{-\sin\sqrt{\text{x}}}{2\sqrt{\text{x}}}$
  2. $\frac{\sin\sqrt{\text{x}}}{2\sqrt{\text{x}}}$
  3. $\sin\sqrt{\text{x}}$
  4. $-\sin\sqrt{\text{x}}$
  1. $7^{\text{x}+\frac{1}{\text{x}}}$
  1. $\Big(\frac{\text{x}^2-1}{\text{x}^2}\Big)\cdot7^{\text{x}+\frac{1}{\text{x}}}\cdot\log7$
  2. $\Big(\frac{\text{x}^2+1}{\text{x}^2}\Big)\cdot7^{\text{x}+\frac{1}{\text{x}}}\cdot\log7$
  3. $\Big(\frac{\text{x}^2-1}{\text{x}^2}\Big)\cdot7^{\text{x}-\frac{1}{\text{x}}}\cdot\log7$
  4. $\Big(\frac{\text{x}^2+1}{\text{x}^2}\Big)\cdot7^{\text{x}-\frac{1}{\text{x}}}\cdot\log7$
  1. $\sqrt\frac{{1-\cos\text{x}}}{1+\cos\text{x}}$
  1. $\frac{1}{2}\sec^2\frac{\text{x}}{2}$
  2. $-\frac{1}{2}\sec^2\frac{\text{x}}{2}$
  3. $\sec^2\frac{\text{x}}{2}$
  4. $-\sec^2\frac{\text{x}}{2}$
  1. $\frac{1}{\text{b}}\tan^{-1}\Big(\frac{\text{x}}{\text{b}}\Big)+\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)$
  1. $\frac{-1}{\text{x}^2+\text{b}^2}+\frac{1}{\text{x}^2+\text{a}^2}$
  2. $\frac{1}{\text{x}^2+\text{b}^2}+\frac{1}{\text{x}^2+\text{a}^2}$
  3. $\frac{1}{\text{x}^2+\text{b}^2}-\frac{1}{\text{x}^2+\text{a}^2}$
  4. None of these.
  1. $\sec^{-1}\text{x}+\text{cosec}^{-1}\frac{\text{x}}{\sqrt{\text{x}^2-1}}$
  1. $\frac{2}{\sqrt{\text{x}^2-1}}$
  2. $\frac{-2}{\sqrt{\text{x}^2-1}}$
  3. $\frac{1}{|\text{x}|\sqrt{\text{x}^2-1}}$
  4. $\frac{2}{|\text{x}|\sqrt{\text{x}^2-1}}$
✓

Answer

  1. (a) $\frac{-\sin\sqrt{\text{x}}}{2\sqrt{\text{x}}}$
Solution:
Let $\text{y}=\cos\sqrt{\text{x}}$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}(\cos\sqrt{\text{x}})=-\sin\sqrt{\text{x}}\cdot\frac{\text{d}}{\text{dx}}(\sqrt{\text{x}})$
$=-\sin\sqrt{\text{x}}\times\frac{1}{2\sqrt{\text{x}}}=\frac{-\sin\sqrt{\text{x}}}{2\sqrt{\text{x}}}$
  1. (a) $\Big(\frac{\text{x}^2-1}{\text{x}^2}\Big)\cdot7^{\text{x}+\frac{1}{\text{x}}}\cdot\log7$
Solution:
Let $\text{y}=7^{\text{x}+\frac{1}{\text{x}}}\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{d}}{\text{dx}}\Big(7^{\text{x}+\frac{1}{\text{x}}}\Big)$
$=7^{\text{x}+\frac{1}{\text{x}}}\cdot\log7\cdot\frac{\text{d}}{\text{dx}}\Big(\text{x}+\frac{1}{\text{x}}\Big)=7^{\text{x}+\frac{1}{\text{x}}}\cdot\log7\cdot\Big(1-\frac{1}{\text{x}^2}\Big)$
$=\Big(\frac{\text{x}^2-1}{\text{x}^2}\Big)\cdot7^{\text{x}+\frac{1}{\text{x}}}\cdot\log7$
  1. (a) $\frac{1}{2}\sec^2\frac{\text{x}}{2}$
Solution:
Let $\text{y}=\sqrt\frac{{1-\cos\text{x}}}{1+\cos\text{x}}=\sqrt{\frac{1-1+2\sin^2\frac{\text{x}}{2}}{2\cos^2\frac{\text{x}}{2}-1+1}}=\tan\Big(\frac{\text{x}}{2}\Big)$
$\therefore\frac{\text{dy}}{\text{dx}}=\sec^2\frac{\text{x}}{2}\cdot\frac{1}{2}=\frac{1}{2}\sec^2\frac{\text{x}}{2}$
  1. (b) $\frac{1}{\text{x}^2+\text{b}^2}+\frac{1}{\text{x}^2+\text{a}^2}$
Solution:
Let $\text{y}=\frac{1}{\text{b}}\tan^{-1}\Big(\frac{\text{x}}{\text{b}}\Big)+\frac{1}{\text{a}}\tan^{-1}\Big(\frac{\text{x}}{\text{a}}\Big)$
$\therefore\frac{\text{dy}}{\text{dx}}=\frac{1}{\text{b}}\times\frac{1}{1+\frac{\text{x}^2}{\text{b}^2}}\times\frac{1}{\text{b}}+\frac{1}{\text{a}}\times\frac{1}{1+\frac{\text{x}^2}{\text{a}^2}}\times\frac{1}{\text{a}}$
$=\frac{1}{\text{x}^2+\text{b}^2}+\frac{1}{\text{x}^2+\text{a}^2}$
  1. (d) $\frac{2}{|\text{x}|\sqrt{\text{x}^2-1}}$
Solution:
Let $\text{y}=\sec^{-1}\text{x}+\text{cosec}^{-1}\frac{\text{x}}{\sqrt{\text{x}^2-1}}$
Put $\text{x}=\sec\theta\Rightarrow\theta=\sec^{-1}\text{x}$
$\therefore\text{y}=\sec^{-1}(\sec\theta)+\text{cosec}^{-1}\Big(\frac{\sec\theta}{\sqrt{\sec^2\theta-1}}\Big)$
$=\theta+\sin^{-1}\Big[\sqrt{1-\cos^2\theta}\Big]$
$=\theta+\sin^{-1}(\sin\theta)=\theta+\theta=2\theta=2\sec^{-1}\text{x}$
$\therefore\frac{\text{dy}}{\text{dx}}=2\frac{\text{d}}{\text{dx}}(\sec^{-1}\text{x})=2\times\frac{1}{|\text{x}|\sqrt{\text{x}^2-1}}$
$=\frac{2}{|\text{x}|\sqrt{\text{x}^2-1}}$

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