Question
Let x = f(t) and y = g(t) be parametric forms with t as a parameter, then
$\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{dt}}\times\frac{\text{dt}}{\text{dx}}=\frac{\text{g}'(\text{t})}{\text{f}'(\text{t})},$ where $\text{f}'(\text{t})\neq0.$
On the basis of above information, answer the following questions.
  1. The derivative of $\text{f}(\tan\text{x})\text{w.r.t.}\text{ g}(\sec\text{x})\text{ at}\text{ x}=\frac{\pi}{4},$ where f'(1) = 2 and $\text{g}'(\sqrt{2})=4,$ is:
  1. $\frac{1}{\sqrt{2}}$
  2. ${\sqrt{2}}$
  3. 1
  4. 0
  1. The derivative of $\sin^{-1}\Big(\frac{2\text{x}}{1+\text{x}^2}\Big)$ with respect to $\cos^{-1}\Big(\frac{1-\text{x}^2}{1+\text{x}^2}\Big)$ is:
  1. -1
  2. 1
  3. 2
  4. 4
  1. The derivative of $\text{e}^{\text{x}^3}$ with respect to log x is:
  1. $\text{e}^{\text{x}^3}$
  2. $3\text{x}^22\text{e}^{\text{x}^3}$
  3. $3\text{x}^3\text{e}^{\text{x}^3}$
  4. $3\text{x}^2\text{e}^{\text{x}^3}+3\text{x}$
  1. The derivative of $\cos^{-1}(2\text{x}^2-1)\text{w.r.t.}\cos^{-1}\text{x}$ is:
  1. $2$
  2. $\frac{-1}{2\sqrt{1-\text{x}^2}}$
  3. $\frac{2}{\text{x}}$
  4. $1-\text{x}^2$
  1. If $\text{y}=\frac{1}{4}\mu^4$ and $\mu=\frac{2}{3}\text{x}^3+5,$ then $\frac{\text{dy}}{\text{dx}}=$
  1. $\frac{2}{27}\text{x}^2(2\text{x}^3+15)^3$
  2. $\frac{2}{7}\text{x}^2(2\text{x}^3+15)^3$
  3. $\frac{2}{27}\text{x}(2\text{x}^3+5)^3$
  4. $\frac{2}{7}(2\text{x}^3+15)^3$

Answer

  1. (a) $\frac{1}{\sqrt{2}}$

Solution:

Now, $\frac{\text{df}(\tan\text{x})}{\text{dg}(\sec\text{x})}=\frac{\text{f}'(\tan \text{x})\sec^2\text{x}}{\text{g}'(\sec\text{x})\sec\text{x}\tan \text{x}}$

$=\frac{\text{f}'(\tan \text{x})\sec\text{x}}{\text{g}'(\sec\text{x})\tan \text{x}}$

$\therefore\Big[\frac{\text{df}(\tan\text{x})}{\text{dg}(\sec\text{x})}\Big]_{\text{x}=\frac{\pi}{4}}=\frac{\text{f}'(1)\sqrt{2}}{\text{g}'(\sqrt{2})\cdot1}=\frac{2\sqrt{2}}{4\cdot1}=\frac{1}{\sqrt{2}}$

  1. (b) 1
  1. (c) $3\text{x}^3\text{e}^{\text{x}^3}$

Solution:

Let $\text{y}=\text{e}^{\text{x}^3},\text{z}=\log\text{x}$

Differentiating w.r.t. x, we get

$\frac{\text{dy}}{\text{dx}}=\text{e}^{\text{x}^3}(3\text{x}^2)=3\text{x}^2\text{e}^{\text{x}^3}$ and $\therefore\frac{\text{dy}}{\text{dz}}=\frac{\frac{\text{dy}}{\text{dx}}}{\frac{\text{dz}}{\text{dx}}}=\frac{3\text{x}^2\text{e}^{\text{x}^3}}{\Big(\frac{1}{\text{x}}\Big)}=3\text{x}^3\text{e}^{\text{x}^3}$

  1. (a) $2$

Solution:

Let $\text{y}=\cos^{-1}(2\text{x}^2-1)=2\cos^{-1}\text{x}$

Differentiating w.r.t. $\cos^{-1}\text{x},$ we get

$\frac{\text{dy}}{\text{d}(\cos^{-1}\text{x})}=\frac{2\text{d}(\cos^{-1}\text{x})}{\text{d}(\cos^{-1}\text{x})}=2$

  1. (a) $\frac{2}{27}\text{x}^2(2\text{x}^3+15)^3$

Solution:

We have, $\text{y}=\frac{1}{4}\text{u}^4\Rightarrow\frac{\text{dy}}{\text{du}}=\frac{1}{4}\cdot4\text{u}^3=\text{u}^3$

and $\text{u}=\frac{2}{3}\text{x}^3+5\Rightarrow\frac{\text{du}}{\text{dx}}=\frac{2}{3}\cdot3\text{x}^2=2\text{x}^2$

$\therefore\frac{\text{dy}}{\text{dx}}=\frac{\text{dy}}{\text{du}}\cdot\frac{\text{du}}{\text{dx}}=\text{u}^3\cdot2\text{x}^2=\Big(\frac{2}{3}\text{x}^3+5\Big)^3(2\text{x})^2$

$=\frac{2}{27}\text{x}^2(2\text{x}^3+15)^3$

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  4. 2
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Image

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For example, every polynomial, constant function are both continuous as well as differentiable and inverse trigonometric functions are continuous and differentiable in its domains etc.
Based on the above information, answer the following questions.
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  3. f(x) is continuous but not differentiable.
  4. None of these.
  1. If $\text{f}(\text{x})=|\text{x}-1|,\text{x }\in\text{ R},$ then at x = 1
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  2. f(x) is continuous but not differentiable.
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  4. None of these.
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  1. If $\text{f}(\text{x})=[\sin\text{x}],$ then which of the following is true?
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Based on the above information, answer the following questions.
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  1. $\begin{bmatrix}1\\-1\end{bmatrix}$
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  1. 2
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