Let $y = sin^{–1} x$

siny =x

Differentiating both sides w.r.t. x, we get

$cos y \frac {dy}{dx}=1$

or

$\frac {dy}{dx} = \frac {1}{cos y} = \frac {1}{cos sin^{–1} x}$

Clearly we can see that x cannot be -1 or 1 as then it will then become undefined. Therefore $x \in (-1,1)$

Also

$cos^2 y=1 -sin^2 y$

$cos^2 y = 1 -x^2$

As $sin^{–1} x$ lies between $(-\pi/2, \pi/2)$, cos y is always positive

Hence

$cos y = \sqrt {1-x^2}$

Therefore

$\frac {dy}{dx} = \frac {1}{\sqrt {1-x^2}}$

Domain of f' is (-1,1)

Let $y = cos^{–1} x$

cosy =x

Differentiating both sides w.r.t. x, we get

$-sin y \frac {dy}{dx}=1$

or

$\frac {dy}{dx} = -\frac {1}{sin y} = -\frac {1}{sin cos^{–1} x}$

Clearly we can see that x cannot be -1 or 1 as then it will then become undefined. Therefore $x \in (-1,1)$

Also

$sin^2 y=1 -cos^2 y$

$sin^2 y = 1 -x^2$

As $cos^{–1} x$ lies between $(0, \pi)$, sin y is always positive

Hence

$sin y = \sqrt {1-x^2}$

Therefore

$\frac {dy}{dx} = -\frac {1}{\sqrt {1-x^2}}$

Domain of f' is (-1,1)

Let $y = tan^{–1} x$

tany =x

Differentiating both sides w.r.t. x, we get

$sec^2 y \frac {dy}{dx}=1$

or

$\frac {dy}{dx} = \frac {1}{sec^2 y} = \frac {1}{1 + tan^2 y}= \frac {1}{1 + x^2}$

Domain of f' is R

Let $y = cot^{–1} x$

cot y =x

Differentiating both sides w.r.t. x, we get

$-cosec^2 y \frac {dy}{dx}=1$

or

$\frac {dy}{dx} = -\frac {1}{cosec^2 y} = \frac {1}{1 + cot^2 y}= -\frac {1}{1 + x^2}=$

Domain of f' is R

Let $y = cosec^{–1} x$

cosec y =x

Differentiating both sides w.r.t. x, we get

$-cosec y cot y \frac {dy}{dx}=1$

or

$\frac {dy}{dx} =- \frac {1}{cosec y cot y} $

Clearly we can see that x cannot be -1 or 1 as then it will then become undefined. Therefore $x \in R -[-1,1]$

Also

$cot^2 y=cosec^2 y -1$

$cot^2 y =x^2 -1 $

As $cosec^{–1} x$ lies between $(-\pi/2, \pi/2)$, cot y will be positive and negative respectively

Hence

When x is negative $cot y = \sqrt {x^2 -1}$

When x is psotive $cot y =- \sqrt {x^2 -1}$

Therefore

When x is negative $\frac {dy}{dx} = -\frac {1}{- x \sqrt {x^2 -1}}=-\frac {1}{ |x| \sqrt {x^2 -1}} $ When is positive $\frac {dy}{dx} = -\frac {1}{ x \sqrt {x^2 -1}}=-\frac {1}{ |x| \sqrt {x^2 -1}} $ Domain of f' is R -[-1,1]

Let $y = sec^{–1} x$

sec y =x

Differentiating both sides w.r.t. x, we get

$sec y tan y \frac {dy}{dx}=1$

or

$\frac {dy}{dx} =\frac {1}{sec y tan y} $

Clearly we can see that x cannot be -1 or 1 as then it will then become undefined. Therefore $x \in R -[-1,1]$

Also

$tan^2 y=sec^2 y -1$

$tan^2 y =x^2 -1 $

As $sec^{–1} x$ lies between $(0, \pi)$, tan y will be positive and negative respectively

Hence

$tan y =\pm \sqrt {x^2 -1}$

Therefore

$\frac {dy}{dx} = \frac {1}{\pm x \sqrt {x^2 -1}} =\frac {1}{ |x| \sqrt {x^2 -1}} $

Domain of f' is R -[-1,1]

Find the derivative of \( y = cos^{-1}(3x) \).

First, recall the derivative of \( \arccos(x) \), which is \( -\frac{1}{\sqrt{1 - x^2}} \).

Then, apply the chain rule:

\[ \frac{dy}{dx} = -\frac{1}{\sqrt{1 - (3x)^2}} \cdot \frac{d}{dx}(3x) \] \[ = -\frac{1}{\sqrt{1 - 9x^2}} \cdot 3 \] \[ = -\frac{3}{\sqrt{1 - 9x^2}} \]

Find the derivative of \( y = x \cdot \arctan(x) \).

Here, we need to use the product rule, which is \( (uv)' = u'v + uv' \), where \( u = x \) and \( v = tan^{-1}(x) \).

Now The derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1 + x^2} \).

So,

\[ \frac{dy}{dx} = 1 \cdot \arctan(x) + x \cdot \frac{1}{1 + x^2} \] \[ = \arctan(x) + \frac{x}{1 + x^2} \]

Find the derivative of \( y = \frac{\sin^{-1}(x)}{x} \) for \( x \neq 0 \).

Apply the quotient rule, \( \left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2} \), where \( u = \sin^{-1}(x) \) and \( v = x \).

Now The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1 - x^2}} \).

So,

\[ \frac{dy}{dx} = \frac{\frac{1}{\sqrt{1 - x^2}} \cdot x - \sin^{-1}(x) \cdot 1}{x^2} \] \[ = \frac{x}{x^2\sqrt{1 - x^2}} - \frac{\sin^{-1}(x)}{x^2} \] \[ = \frac{1}{x\sqrt{1 - x^2}} - \frac{\sin^{-1}(x)}{x^2} \]

**Notes**-
**NCERT Solutions & Assignments**- NCERT Solution for Continuity And differentiability Exercise 5.1
- NCERT Solution for Continuity And differentiability Exercise 5.2
- NCERT Solution for Continuity And differentiability Exercise 5.3
- NCERT Solution for Continuity And differentiability Exercise 5.4
- Continuity and Differentiability class 12 Important questions

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