Class 12 RD Sharma Solutions – Chapter 11 Differentiation – Exercise 11.2 | Set 1

Last Updated : 14 Jul, 2021

Question 1. Differentiate y = sin (3x + 5) with respect to x.

Solution:

We have,
y = sin (3x + 5)
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\sin\left( 3x + 5 \right)$
On using chain rule, we have
$\frac{d y}{d x} = \cos\left( 3x + 5 \right)\frac{d}{dx}\left( 3x + 5 \right)$
$\frac{d y}{d x} = \cos\left( 3x + 5 \right) \times 3$
$\frac{d y}{d x} = 3\cos\left( 3x + 5 \right)$

Question 2. Differentiate y = tan² x with respect to x.

Solution:

We have,
y = tan² x
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}(\tan^2 x)$
On using chain rule, we have
$\frac{d y}{d x} = 2 \tan x\frac{d}{dx}\left( \tan x \right)$
$\frac{d y}{d x} = 2 \tan x \times \sec^2 x$
$\frac{d y}{d x} = 2 \tan x\sec^2 x$

Question 3. Differentiate y = tan (x + 45°) with respect to x.

Solution:

We have,
y = tan (x + 45°)
y = $\tan\left\{ \left( x + 45 \right)\frac{\pi}{180} \right\}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\tan\left\{ \left( x + 45 \right)\frac{\pi}{180} \right\}$
On using chain rule, we have
$\frac{d y}{d x} = \sec^2 \left\{ \left( x + 45 \right)\frac{\pi}{180} \right\} \times \frac{d}{dx}\left( x + 45 \right)\frac{\pi}{180}$
$\frac{d y}{d x} = \frac{\pi}{180} \sec^2 \left( x^\circ + 45^\circ \right)$

Question 4. Differentiate y = sin (log x) with respect to x.

Solution:

We have,
y = sin (log x)
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\sin\left( \log x \right)$
On using chain rule, we have
$\frac{d y}{d x} = \cos\left( \log x \right)\frac{d}{dx}\left( \log x \right)$
$\frac{d y}{d x} = \frac{1}{x}\cos\left( \log x \right)$

Question 5. Differentiate y = e^{sin √x} with respect to x.

Solution:

We have,
y = e^{sin √x}
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( e^{\sin \sqrt{x}} \right)$
On using chain rule, we have
$\frac{d y}{d x} = e^{\sin \sqrt{x}} \frac{d}{dx}\left( \sin\sqrt{x} \right)$
On using chain rule again, we have
$\frac{d y}{d x} = e^{\sin \sqrt{x}} \times \cos\sqrt{x}\frac{d}{dx}\sqrt{x}$
$\frac{d y}{d x} = e^{\sin \sqrt{x}} \times \cos\sqrt{x} \times \frac{1}{2\sqrt{x}}$
$\frac{d y}{d x} = \frac{\cos\sqrt{x} e^{\sin}\sqrt{x}}{2\sqrt{x}}$

Question 6. Differentiate y = e^{tan x} with respect to x.

Solution:

We have,
y = e^{tan x}
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( e^{\tan x} \right)$
On using chain rule, we have
$\frac{d y}{d x} = e^{\tan x} \frac{d}{dx}\left( \tan x \right)$
$\frac{d y}{d x} = e^{\tan x}\sec^2 x$

Question 7. Differentiate y = sin² (2x + 1) with respect to x.

Solution:

We have,
y = sin² (2x + 1)
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left[ \sin^2 \left( 2x + 1 \right) \right]$
On using chain rule, we have
$\frac{d y}{d x} = 2\sin\left( 2x + 1 \right)\frac{d}{dx}\sin\left( 2x + 1 \right)$
On using chain rule again, we have
$\frac{d y}{d x} = 2\sin\left( 2x + 1 \right) \cos\left( 2x + 1 \right) \frac{d}{dx}\left( 2x + 1 \right)$
$\frac{d y}{d x} = 4\sin\left( 2x + 1 \right) \cos\left( 2x + 1 \right)$
As sin 2A = 2 sin A cos A, we get
$\frac{d y}{d x} = 2\sin2\left( 2x + 1 \right)$
$\frac{d y}{d x} = 2 \sin\left( 4x + 2 \right)$

Question 8. Differentiate y = log₇ (2x − 3) with respect to x.

Solution:

We have,
y = log₇ (2x − 3)
As $\log_a b = \frac{\log b}{\log a}$ , we have
y = $\frac{\log\left( 2x - 3 \right)}{\log7}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{1}{\log7}\frac{d}{dx}\left\{ \log\left( 2x - 3 \right) \right\}$
On using chain rule, we have
$\frac{d y}{d x} = \frac{1}{\log7} \times \frac{1}{\left( 2x - 3 \right)}\frac{d}{dx}\left( 2x - 3 \right)$
$\frac{d y}{d x} = \frac{2}{\left( 2x - 3 \right)\log7}$

Question 9. Differentiate y = tan 5x° with respect to x.

Solution:

We have,
y = tan 5x°
y = $\tan\left( 5x \times \frac{\pi}{180} \right)$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\tan\left( 5x \times \frac{\pi}{180} \right)$
On using chain rule, we have
$\frac{d y}{d x} = \sec^2 \left( 5x \times \frac{\pi}{180} \right)\frac{d}{dx}\left( 5x \times \frac{\pi}{180} \right)$
$\frac{d y}{d x} = \left( \frac{5\pi}{180} \right) \sec^2 \left( 5x \times \frac{\pi}{180} \right)$
$\frac{d y}{d x} = \frac{5\pi}{180} \sec^2 \left( 5x^\circ\right)$

Question 10. Differentiate y = $2^{x^3}$ with respect to x.

Solution:

We have,
y = $2^{x^3}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( 2^{x^3} \right)$
On using chain rule, we have
$\frac{d y}{d x} = 2^{x^3} \times \log_e 2\frac{d}{dx}\left( x^3 \right)$
$\frac{d y}{d x} = 3 x^2 \times 2^{x^3} \times \log_e 2$

Question 11. Differentiate y = $3^{e^x}$ with respect to x.

Solution:

We have,
y = $3^{e^x}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( 3^{e^x} \right)$
On using chain rule, we have
$\frac{d y}{d x} = 3^{e^x} \log3\frac{d}{dx}\left( e^x \right)$
$\frac{d y}{d x} = e^x \times 3^{e^x} \log3$

Question 12. Differentiate y = log_x 3 with respect to x.

Solution:

We have,
y = log_x 3
As $\log_a b = \frac{\log b}{\log a}$ , we get
y = $\frac{\log3}{\log x}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( \frac{\log3}{\log x} \right)$
$\frac{d y}{d x} = \log3\frac{d}{dx} \left( \log x \right)^{- 1}$
On using chain rule, we have
$\frac{d y}{d x} = \log3 \times \left[ - 1 \left( \log x \right)^{- 2} \right]\frac{d}{dx}\left( \log x \right)$
$\frac{d y}{d x} = - \frac{\log3}{\left( \log x \right)^2} \times \frac{1}{x}$
$\frac{d y}{d x} = - \left( \frac{\log3}{\log x} \right)^2 \times \frac{1}{x} \times \frac{1}{\log3}$
As $\frac{\log b}{\log a} = \log_a b$ , we get
$\frac{d y}{d x} = - \frac{1}{x\log3 \left( \log_3 x \right)^2}$

Question 13. Differentiate y = $3^{x^2 + 2x}$ with respect to x.

Solution:

We have,
y = $3^{x^2 + 2x}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( 3^{x^2 + 2x} \right)$
On using chain rule, we have
$\frac{d y}{d x} = 3^{x^2 + 2x} \times \log_e 3\frac{d}{dx}\left( x^2 + 2x \right)$
$\frac{d y}{d x} = \left( 2x + 2 \right) 3^{x^2 + 2x} \log_e 3$

Question 14. Differentiate y = $\sqrt{\frac{a^2 - x^2}{a^2 + x^2}}$ with respect to x.

Solution:

We have,
y = $\sqrt{\frac{a^2 - x^2}{a^2 + x^2}}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( \sqrt{\frac{a^2 - x^2}{a^2 + x^2}} \right)$
On using chain rule, we have
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^{\frac{1}{2} - 1} \times \frac{d}{dx}\left( \frac{a^2 - x^2}{a^2 + x^2} \right)$
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{a^2 - x^2}{a^2 + x^2} \right)^\frac{- 1}{2} \times \left\{ \frac{\left( a^2 + x^2 \right)\frac{d}{dx}\left( a^2 - x^2 \right) - \left( a^2 - x^2 \right)\frac{d}{dx}\left( a^2 + x^2 \right)}{\left( a^2 + x^2 \right)^2} \right\}$
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{a^2 + x^2}{a^2 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 2x\left( a^2 + x^2 \right) - 2x\left( a^2 - x^2 \right)}{\left( a^2 + x^2 \right)^2} \right\}$
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{a^2 + x^2}{a^2 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 2x a^2 - 2 x^3 - 2x a^2 + 2 x^3}{\left( a^2 + x^2 \right)^2} \right\}$
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{a^2 + x^2}{a^2 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 4x a^2}{\left( a^2 + x^2 \right)^2} \right\}$
$\frac{d y}{d x} = \frac{- 2x a^2}{\sqrt{a^2 - x^2} \left( a^2 + x^2 \right)^\frac{3}{2}}$

Question 15. Differentiate y = $3^{x \log x}$ with respect to x.

Solution:

We have,
y = $3^{x \log x}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left( 3^{x \log x} \right)$
On using chain rule, we have
$\frac{d y}{d x} = 3^{x \log x} \times \log_e 3\frac{d}{dx}\left( x \log x \right)$
$\frac{d y}{d x} = 3^x \log x \times \log_e 3\left[ x\frac{d}{dx}\left( \log x \right) + \log x\frac{d}{dx}\left( x \right) \right]$
$\frac{d y}{d x} = 3^{x \log x} \times \log_e 3\left[ \frac{x}{x} + \log x \right]$
$\frac{d y}{d x} = 3^{x \log x} \left( 1 + \log x \right) \times \log_e 3$
$\frac{d y}{d x} = 3^{x \log x} \left( 1 + \log x \right)\log_e 3$

Question 16. Differentiate y = $\sqrt{\frac{1 + \sin x}{1 - \sin x}}$ with respect to x.

Solution:

We have,
y = $\sqrt{\frac{1 + \sin x}{1 - \sin x}}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx} \left( \sqrt{\frac{1 + \sin x}{1 - \sin x}} \right)$
$\frac{d y}{d x} = \frac{d}{dx} \left( \frac{1 + \sin x}{1 - \sin x} \right)^\frac{1}{2}$
On using chain rule, we have
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 + \sin x}{1 - \sin x} \right)^{\frac{1}{2} - 1} \frac{d}{dx}\left( \frac{1 + \sin x}{1 - \sin x} \right)$
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 - \sin x}{1 + \sin x} \right)^\frac{1}{2} \left[ \frac{\left( 1 - \sin x \right)\left( \cos x \right) - \left( 1 + \sin x \right)\left( - \cos x \right)}{\left( 1 - \sin x \right)^2} \right]$
$\frac{d y}{d x} = \frac{1}{2}\frac{\left( 1 - \sin x \right)^\frac{1}{2}}{\left( 1 + \sin x \right)^\frac{1}{2}}\left[ \frac{\cos x - \cos x \sin x + \cos x + \sin x \cos x}{\left( 1 - \sin x \right)^2} \right]$
$\frac{d y}{d x} = \frac{1}{2} \times \frac{2\cos x}{\sqrt{1 + \sin x}\left( 1 - \sin x \right)\frac{3}{2}}$
$\frac{d y}{d x} = \frac{\cos x}{\sqrt{1 + \sin x}\left( 1 - \sin x \right)\frac{3}{2}}$
$\frac{d y}{d x} = \frac{\cos x}{\sqrt{1 + \sin x}\sqrt{1 - \sin x}\left( 1 - \sin x \right)}$
$\frac{d y}{d x} = \frac{\cos x}{\sqrt{1 - \sin^2 x} \times \left( 1 - \sin x \right)}$
$\frac{d y}{d x} = \frac{\cos x}{\cos x\left( 1 - \sin x \right)}$
$\frac{d y}{d x} = \frac{1}{\left( 1 - \sin x \right)} \times \frac{\left( 1 + \sin x \right)}{\left( 1 + \sin x \right)}$
$\frac{d y}{d x} = \frac{\left( 1 + \sin x \right)}{\left( 1 - \sin^2 x \right)}$
$\frac{d y}{d x} = \frac{1 + \sin x}{\cos^2 x}$
$\frac{d y}{d x} = \frac{1}{\cos x}\left( \frac{1}{\cos x} + \frac{\sin x}{\cos x} \right)$
$\frac{d y}{d x} = \sec x\left( \sec x + \tan x \right)$

Question 17. Differentiate y = $\sqrt{\frac{1 - x^2}{1 + x^2}}$ with respect to x.

Solution:

We have,
y = $\sqrt{\frac{1 - x^2}{1 + x^2}}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx} \left( \sqrt{\frac{1 - x^2}{1 + x^2}} \right)$
$\frac{d y}{d x} = \frac{d}{dx} \left( \frac{1 - x^2}{1 + x^2} \right)^\frac{1}{2}$
On using chain rule, we have
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 - x^2}{1 + x^2} \right)^{\frac{1}{2} - 1} \times \frac{d}{dx}\left( \frac{1 - x^2}{1 + x^2} \right)$
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 - x^2}{1 + x^2} \right)^\frac{- 1}{2} \times \left\{ \frac{\left( 1 + x^2 \right)\frac{d}{dx}\left( 1 - x^2 \right) - \left( 1 - x^2 \right)\frac{d}{dx}\left( 1 + x^2 \right)}{\left( 1 + x^2 \right)^2} \right\}$
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 + x^2}{1 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 2x\left( 1 + x^2 \right) - 2x\left( 1 - x^2 \right)}{\left( 1 + x^2 \right)^2} \right\}$
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 + x^2}{1 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 2x - 2 x^3 - 2x + 2 x^3}{\left( 1 + x^2 \right)^2} \right\}$
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 + x^2}{1 - x^2} \right)^\frac{1}{2} \left\{ \frac{- 4x}{\left( 1 + x^2 \right)^2} \right\}$
$\frac{d y}{d x} = \frac{- 2x}{\sqrt{1 - x^2} \left( 1 + x^2 \right)^\frac{3}{2}}$

Question 18. Differentiate y = (log sin x)² with respect to x.

Solution:

We have,
y = (log sin x)²
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx} \left( \log \sin x \right)^2$
On using chain rule, we have
$\frac{d y}{d x} = 2\left( \log \sin x \right)\frac{d}{dx}\left( \log \sin x \right)$
$\frac{d y}{d x} = 2\left( \log \sin x \right) \times \frac{1}{\sin x}\frac{d}{dx}\left( \sin x \right)$
$\frac{d y}{d x} = 2\left( \log \sin x \right) \times \frac{1}{\sin x} \times \cos x$
$\frac{d y}{d x} = 2\left( \log \sin x \right)\cot x$

Question 19. Differentiate y = $\sqrt{\frac{1 + x}{1 - x}}$ with respect to x.

Solution:

We have,
y = $\sqrt{\frac{1 + x}{1 - x}}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx} \left( \sqrt{\frac{1 + x}{1 - x}} \right)$
$\frac{d y}{d x} = \frac{d}{dx} \left( \frac{1 + x}{1 - x} \right)^\frac{1}{2}$
On using chain rule, we have
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 + x}{1 - x} \right)^{\frac{1}{2} - 1} \times \frac{d}{dx}\left( \frac{1 + x}{1 - x} \right)$
On using quotient rule, we have
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 + x}{1 - x} \right)^\frac{- 1}{2} \times \left\{ \frac{\left( 1 - x \right)\frac{d}{dx}\left( 1 + x \right) - \left( 1 + x \right)\frac{d}{dx}\left( 1 - x \right)}{\left( 1 - x \right)^2} \right\}$
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 - x}{1 + x} \right)^\frac{1}{2} \left\{ \frac{\left( 1 - x \right)\left( 1 \right) - \left( 1 + x \right)\left( - 1 \right)}{\left( 1 - x \right)^2} \right\}$
$\frac{d y}{d x} = \frac{1}{2} \left( \frac{1 - x}{1 + x} \right)^\frac{1}{2} \left\{ \frac{1 - x + 1 + x}{\left( 1 - x \right)^2} \right\}$
$\frac{d y}{d x} = \frac{1}{2}\frac{\left( 1 - x \right)^\frac{1}{2}}{\left( 1 + x \right)^\frac{1}{2}} \times \frac{2}{\left( 1 - x \right)^2}$
$\frac{d y}{d x} = \frac{1}{\sqrt{1 + x} \left( 1 - x \right)^\frac{3}{2}}$

Question 20. Differentiate y = $\sin \left( \frac{1 + x^2}{1 - x^2} \right)$ with respect to x.

Solution:

We have,
y = $\sin \left( \frac{1 + x^2}{1 - x^2} \right)$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx} \left( \sin \left( \frac{1 + x^2}{1 - x^2} \right) \right)$
On using chain rule, we have
$\frac{d y}{d x} = \cos x\left( \frac{1 + x^2}{1 - x^2} \right)\frac{d}{dx}\left( \frac{1 + x^2}{1 - x^2} \right)$
On using quotient rule, we have
$\frac{d y}{d x} = \cos x\left( \frac{1 + x^2}{1 - x^2} \right)\left[ \frac{\left( 1 - x^2 \right)\frac{d}{dx}\left( 1 + x^2 \right) - \left( 1 + x^2 \right)\frac{d}{dx}\left( 1 - x^2 \right)}{\left( 1 - x^2 \right)^2} \right]$
$\frac{d y}{d x} = \cos x\left( \frac{1 + x^2}{1 - x^2} \right)\left[ \frac{\left( 1 - x^2 \right)\left( 2x \right) - \left( 1 + x^2 \right)\left( - 2x \right)}{\left( 1 - x^2 \right)^2} \right]$
$\frac{d y}{d x} = \cos x\left( \frac{1 + x^2}{1 - x^2} \right)\left[ \frac{2x - 2 x^3 + 2x + 2 x^3}{\left( 1 - x^2 \right)^2} \right]$
$\frac{d y}{d x} = \frac{4x}{\left( 1 - x^2 \right)^2}\cos x\left( \frac{1 + x^2}{1 - x^2} \right)$

Question 21. Differentiate y = $e^{3x} \cos2x$ with respect to x.

Solution:

We have,
y = $e^{3x} \cos2x$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx} \left( e^{3x} \cos2x\right)$
On using product rule, we have
$\frac{d y}{d x} = e^{3x} \times \frac{d}{dx}\left( \cos2x \right) + \cos2x\frac{d}{dx}\left( e^{3x} \right)$
On using chain rule, we have
$\frac{d y}{d x} = e^{3x} \times \left( - \sin2x \right)\frac{d}{dx}\left( 2x \right) + \cos2x e^{3x} \frac{d}{dx}\left( 3x \right)$
$\frac{d y}{d x} = - 2 e^{3x} \sin2x + 3 e^{3x} \cos2x$
$\frac{d y}{d x} = e^{3x} \left( 3 \cos2x - 2 \sin2x \right)$

Question 22. Differentiate y = sin(log sin x) with respect to x.

Solution:

We have,
y = sin(log sin x)
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx} \left(sin(log sin x)\right)$
On using chain rule, we have
$\frac{d y}{d x}=\cos(\log \sin x)\frac{d}{dx}(\log \sin x)$
On using chain rule again, we have
$\frac{d y}{d x}=\cos (\log \sin x)\frac{1}{sin x}\frac{d}{dx}(\sin x)$
$\frac{d y}{d x}=\cos (\log \sin x)\frac{cos x}{sin x}$
$\frac{d y}{d x}=\cos (\log \sin x) \cot x$

Question 23. Differentiate y = e^{tan 3x} with respect to x.

Solution:

We have,
y = e^{tan 3x}
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left(e^{\tan3 x} \right)$
On using chain rule, we have
$\frac{d y}{d x} = e^{\tan3x} \frac{d}{dx}\left( \tan3x \right)$
$\frac{d y}{d x} = e^{\tan3x} \sec^2 3x \times \frac{d}{dx}\left( 3x \right)$
$\frac{d y}{d x} = e^{\tan3x} \sec^2 3x \times 3$
$\frac{d y}{d x} = 3e^{\tan3x} \sec^2 3x$

Question 24. Differentiate y = $e^{\sqrt{\cot x}}$ with respect to x.

Solution:

We have,
y = $e^{\sqrt{\cot x}}$
On differentiating y with respect to x we get,
$\frac{d y}{d x} = \frac{d}{dx}\left(e^{\sqrt{\cot x}} \right)$
$\frac{d y}{d x} = \frac{d}{dx}\left( e^{\left( \cot x \right)^\frac{1}{2} }\right)$
On using chain rule, we have
$\frac{d y}{d x} = e^{\left( \cot x \right)^\frac{1}{2}} \times \frac{d}{dx} \left( \cot x \right)^\frac{1}{2}$
$\frac{d y}{d x} = e^{\sqrt{\cot x}} \times \frac{1}{2} \left( \cot x \right)^{\frac{1}{2} - 1} \frac{d}{dx}\left( \cot x \right)$
$\frac{d y}{d x} = - \frac{e^{\sqrt{\cot x}}{cosec}^2 x}{2\sqrt{\cot x}}$