1. ⇒ (MHT CET 2023 12th May Morning Shift )

The derivative of $f (\tan x)$ w.r.t. $g (\sec x)$ at $x = \frac{π}{4}$ where $f^{'} (1) = 2$ and $g^{'} (\sqrt{2}) = 4$ is

A. $\frac{1}{\sqrt{2}}$

B. $\sqrt{2}$

C. 1

D. 0

Correct Option is (A)

$\begin{aligned} Let p = f (\tan x) and q = g (\sec x) \\ ∴ \frac{dp}{d x} = f^{'} (\tan x) \times \sec^{2} x and \\ \frac{dq}{d x} = g^{'} (\sec x) \times \sec x \tan x \\ {∴ \frac{dp}{dx} |}_{x = \frac{π}{4}} = f^{'} (1) \times 2 = 4, \\ {\frac{dq}{d x} |}_{x = \frac{π}{4}} = g^{'} (\sqrt{2}) \times \sqrt{2} = 4 \sqrt{2} \\ ∴ Required Derivative = (\frac{{\frac{dp}{d x} |}_{x = \frac{π}{4}}}{{\frac{dq}{d x} |}_{x = \frac{π}{4}}}) = \frac{4}{4 \sqrt{2}} = \frac{1}{\sqrt{2}} \end{aligned}$

2. ⇒ (MHT CET 2023 11th May Morning Shift )

Derivative of $\tan^{- 1} (\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}})$ w.r.t. $\cos^{- 1} x^{2}$ is

A. $- \frac{1}{2}$

B. $- 1$

C. $\frac{1}{2}$

D. 1

Correct Option is (A)

Let $y = \tan^{- 1} (\frac{\sqrt{1 + x^{2}} - \sqrt{1 - x^{2}}}{\sqrt{1 + x^{2}} + \sqrt{1 - x^{2}}})$ and $z = \cos^{- 1} (x^{2})$

Put $x^{2} = \cos 2 θ \Rightarrow θ = \frac{1}{2} \cos^{- 1} x^{2}$

$\begin{aligned} ∴ y = \tan^{- 1} (\frac{\sqrt{1 + \cos 2 θ} - \sqrt{1 - \cos 2 θ}}{\sqrt{1 + \cos 2 θ} + \sqrt{1 - \cos 2 θ}}) \\ \Rightarrow y = \tan^{- 1} (\frac{\cos θ - \sin θ}{\cos θ + \sin θ}) \\ \Rightarrow y = \tan^{- 1} (\frac{1 - \tan θ}{1 + \tan θ}) \\ \Rightarrow y = \tan^{- 1} (\tan (\frac{π}{4} - θ)) \\ \Rightarrow y = \frac{π}{4} - \frac{1}{2} \cos^{- 1} x^{2} \\ \Rightarrow y = \frac{π}{4} - \frac{1}{2} z \\ ∴ \frac{dy}{dz} = - \frac{1}{2} \end{aligned}$

3. ⇒ (MHT CET 2023 9th May Morning Shift )

The rate of change of $\sqrt{x^{2} + 16}$ with respect to $\frac{x}{x - 1}$ at $x = 5$ is

A. $\frac{- 80}{\sqrt{41}}$

B. $\frac{80}{\sqrt{41}}$

C. $\frac{12}{5}$

D. $\frac{- 12}{5}$

Correct Option is (A)

$\begin{array}{ll} y = \sqrt{x^{2} + 16} \\ ∴ \frac{d y}{d x} = \frac{2 x}{2 \sqrt{x^{2} + 16}} \\ ∴ \frac{d y}{d x} = \frac{x}{\sqrt{x^{2} + 16}} ..... (i) \\ Let z = \frac{x}{x - 1} \\ ∴ \frac{dz}{d x} = \frac{(x - 1) - x}{(x - 1)^{2}} \\ ∴ \frac{dz}{d x} = \frac{- 1}{(x - 1)^{2}} ..... (ii) \\ ∴ {(\frac{d y}{dz})}_{x = 5} = \frac{\frac{x}{\sqrt{x^{2} + 16}}}{\frac{- 1}{(x - 1)^{2}}} \\ = \frac{- 5}{\sqrt{25 + 16}} \times 16 = \frac{- 80}{\sqrt{41}} \end{array}$

4. ⇒ (MHT CET 2021 21th September Morning Shift )

The derivative of $(\log x)^{x}$ with respect to $\log x$ is

A. $(\log x)^{x} [\frac{1}{\log x} \log (\log x)]$

B. $(\log x)^{x} [\log x + \frac{1}{\log (\log x)}]$

C. $x (\log x)^{x} [\frac{1}{\log x} + \log (\log x)]$

D. $x (\log x)^{x} [\log x + \frac{1}{\log (\log x)}]$

Correct Option is (C)

$\begin{aligned} Let = (\log x)^{x} \\ ∴ \log u = x \log [\log (x)] \\ ∴ \frac{1}{u} \frac{d u}{d x} = \frac{x}{\log (x)} \times \frac{1}{x} + \log (\log x) = \log (\log x) + \frac{1}{\log x} \\ ∴ \frac{d u}{d x} = (\log x)^{x} [\frac{1}{\log x} + \log (\log x)] .....(1) \\ Let v = \log x \Rightarrow \frac{d v}{d x} = \frac{1}{x} .....(2) \\ ∴ \frac{d u}{d v} = (\log x)^{x} (x) [\frac{1}{\log x} + \log (\log x)] .....(From (1), (2)) \end{aligned}$

⇒Derivative