Informative line

### Domain And Derivative Of Inverse Trigonometric Functions

Learn how to find domain and range of tangent, trig, sine inverse function. Practice derivative of inverse cosine and tangent, Sine.

# Domain and Range of Inverse Sine Functions

• Domain of $$sin^{-1}x$$ is $$-1\leq x\leq 1$$ or $$[-1,1]$$
• Range of $$sin^{-1}x$$ is $$-\dfrac {\pi}{2}\leq y\leq \dfrac {\pi}{2}$$ or $$\left [-\dfrac {\pi}{2},\;\dfrac {\pi}{2}\right]$$
• To find the domain of $$sin^{-1} (g(x))$$, where  $$g(x)$$ is any expression we solve $$-1\leq g(x)\leq 1$$

The graph of  $$sin^{-1}x$$ v/s $$x$$ is as shown below.

#### The domain of  $$f(x) = sin^{-1}(2x-3)$$  is

A $$x\in[1, 2]$$

B $$x\in(-\infty, 4]$$

C $$x\in[-1, 3]$$

D $$x\in[3, \infty)$$

×

Domain of $$sin^{-1}(g(x))$$ is $$-1\leq g(x) \leq 1$$.

$$\therefore$$  Domain of $$sin^{-1}(2x-3)$$ is $$-1\leq 2x-3 \leq 1$$.

$$\Rightarrow 2\leq 2x\leq 4$$

$$\Rightarrow 1\leq x\leq 2$$

$$=x\in[1, 2]$$

### The domain of  $$f(x) = sin^{-1}(2x-3)$$  is

A

$$x\in[1, 2]$$

.

B

$$x\in(-\infty, 4]$$

C

$$x\in[-1, 3]$$

D

$$x\in[3, \infty)$$

Option A is Correct

# Derivative of Inverse Sine Function

• $$\dfrac {d}{dx}(sin^{-1}x)=\dfrac {1}{\sqrt {1-x^2}} (-1 \leq x\leq 1)$$?
• $$\dfrac {d}{dx}\Big(sin^{-1}f(x)\Big)=\underbrace{\dfrac {1}{\sqrt {1-(f(x))^2}}×f'(x)}_{chain\,rule}$$

#### If $$f(x)=sin^{-1}\Big( \sqrt {cos\,x}\Big)$$, find $$f'(x)$$.

A $$\dfrac {-cos\,x}{2\sqrt {sin\,x}\sqrt {1-cos\,x}}$$

B $$\dfrac {1}{\sqrt {1-sin\,x}}$$

C $$\dfrac {-sin\,x}{2\sqrt {cos\,x}\sqrt {1-sin\,x}}$$

D $$cos^2\,x-sin\,x+C$$

×

$$f(x)=sin^{-1}\Big( \sqrt {cos\,x}\Big)$$

$$\Rightarrow f'(x)=\dfrac {1}{\sqrt {1-\Big( \sqrt {cos\,x}\Big)^2}}×\dfrac {d}{dx}\Big(\sqrt {cos\,x}\Big)$$

$$=\dfrac {1}{\sqrt {1-sin\,x}}×\dfrac {d}{dx}(cos\,x)^{1/2}$$

$$=\dfrac {1}{\sqrt {1-sin\,x}}×\dfrac {1}{2}(cos\,x)^{-1/2}×(-sin\,x)$$

$$=\dfrac {-sin\,x}{2\sqrt {cos\,x}\sqrt {1-sin\,x}}$$

### If $$f(x)=sin^{-1}\Big( \sqrt {cos\,x}\Big)$$, find $$f'(x)$$.

A

$$\dfrac {-cos\,x}{2\sqrt {sin\,x}\sqrt {1-cos\,x}}$$

.

B

$$\dfrac {1}{\sqrt {1-sin\,x}}$$

C

$$\dfrac {-sin\,x}{2\sqrt {cos\,x}\sqrt {1-sin\,x}}$$

D

$$cos^2\,x-sin\,x+C$$

Option C is Correct

# Derivative of Inverse Tangent Function

• $$\dfrac {d}{dx}\left ( tan^{-1}\,x \right)=\dfrac {1}{1+x^2}$$
• $$\dfrac {d}{dx}\left ( tan^{-1}\,\Big(f(x)\Big) \right)$$

$$=\dfrac {1}{1+(f(x))^2}×f'(x)$$

$$=\dfrac {f'(x)}{1+(f(x))^2}$$

#### If $$f(x)=tan^{-1}(e^{3x})$$, then $$f'(x)$$ is

A $$\dfrac {sin\,x}{e^x}$$

B $$\dfrac {e^{3x}}{1+e^{6x}}$$

C $$\dfrac {e^{5x}}{x+4}$$

D $$\dfrac {3e^{3x}}{1+e^{6x}}$$

×

$$f(x)=tan^{-1}(e^{3x})$$

$$\Rightarrow \;f'(x)=\dfrac {1}{1+(e^{3x})^2}×\dfrac {d}{dx}(e^{3x})$$

$$f'(x)=\dfrac {1}{1+(e^{6x})}×(e^{3x})×3=\dfrac {3\,e^{3x}}{1+e^{6x}}$$

### If $$f(x)=tan^{-1}(e^{3x})$$, then $$f'(x)$$ is

A

$$\dfrac {sin\,x}{e^x}$$

.

B

$$\dfrac {e^{3x}}{1+e^{6x}}$$

C

$$\dfrac {e^{5x}}{x+4}$$

D

$$\dfrac {3e^{3x}}{1+e^{6x}}$$

Option D is Correct

# Domain and Range of Tangent Inverse Function

• Consider $$f(x)=tan\,x$$
• If we apply the horizontal line test we see that $$f$$ is not one-to-one.

• Now if we restrict the domain to  $$\left (\dfrac {-\pi}{2}\,,\dfrac {\pi}{2} \right)$$ i.e. $$\dfrac {-\pi}{2}\,<x<\dfrac {\pi}{2}$$then we see that  $$tan\,x$$  has become one-to-one.

The inverse function of this restricted tangent function is called inverse tangent function or $$tan^{-1}\,x$$ or arc$$(tan\,x)$$.

• If $$f(x)=tan\,x$$ then  $$f^{-1}(x)=tan^{-1}\,x$$ or arc $$(tan\,x)$$
• If $$y=tan^{-1}\,x\;\Rightarrow x = tan\,y$$    $$\left (\dfrac {-\pi}{2}\,<y<\dfrac {\pi}{2}\right)$$
• If $$x\in R$$ then $$tan^{-1}\,x$$ is the angle between $$\dfrac {-\pi}{2}$$ and $$\dfrac {\pi}{2}$$ whose tangent is $$x$$.
• If $$tan^{-1}\,(-1)$$ is the angle between $$\dfrac {-\pi}{2}$$ and $$\dfrac {\pi}{2}$$ whose tangent is $$-1=\dfrac {-\pi}{4}$$.

$$\therefore \;\; tan^{-1}(-1)=\dfrac {-\pi}{4}$$

• The domain of $$f(x) = tan ^{-1}x$$ is $$-\infty<x<\infty$$
• The range of  $$y=f(x)=tan^{-1}x$$ is $$\left ( \dfrac {-\pi}{2}<y<\;\dfrac {\pi}{2} \right)$$

#### The Range of $$y=tan^{-1}\,(5x-2)$$ is

A $$\left [ \dfrac {-\pi}{2},\;\dfrac {\pi}{2} \right]$$

B $$\left ( \dfrac {-\pi}{2},\;\dfrac {\pi}{2} \right)$$

C $$(-\infty, \;\infty)$$

D $$\left ( \dfrac {-5\pi}{2},\;\dfrac {5\pi}{2} \right)$$

×

Function, $$y=tan^{-1}\,(5x-2)$$

$$tan\,y=5x-2$$

$$\forall\,x\in R \Rightarrow y\in \left ( \dfrac {-\pi}{2},\;\dfrac {\pi}{2} \right)$$

$$\therefore$$ The range of $$f(x)=tan^{-1}\,(5x-2)$$ is $$\left ( \dfrac {-\pi}{2},\;\dfrac {\pi}{2} \right)$$.

### The Range of $$y=tan^{-1}\,(5x-2)$$ is

A

$$\left [ \dfrac {-\pi}{2},\;\dfrac {\pi}{2} \right]$$

.

B

$$\left ( \dfrac {-\pi}{2},\;\dfrac {\pi}{2} \right)$$

C

$$(-\infty, \;\infty)$$

D

$$\left ( \dfrac {-5\pi}{2},\;\dfrac {5\pi}{2} \right)$$

Option B is Correct

# Domain and Range of Cosine Inverse Function

• The domain of $$f(x) = cos^{-1}x$$ is $$[-1,1]$$ or $$-1 \leq x \leq 1$$
• The range of $$f(x) = cos^{-1}x$$ is $$[0,\pi]$$ or $$0 \leq f(x) \leq \pi$$.
• The domain of  $$cos^{-1}(g(x))$$ is  obtained by solving $$-1\leq g(x) \leq 1$$.

#### The domain of $$f(x) = cos^{-1}(3x-2)$$ is

A $$x\in\left [ \dfrac {5}{4},\;7 \right]$$

B $$x\in\left [ \dfrac {-2}{3},\;\dfrac {1}{3} \right]$$

C $$x\in\left [ \dfrac {1}{3},\;1 \right]$$

D $$x\in\left [ 2,\;\infty \right)$$

×

Domain of $$cos^{-1}(g(x))$$ is   $$-1\leq g(x) \leq 1$$.

$$\therefore$$  Domain of $$cos^{-1}(3x-2)$$ is $$-1\leq 3x-2 \leq 1$$

$$\Rightarrow 1\leq 3x \leq 3$$

$$\Rightarrow\dfrac {1}{3}\leq x \leq 1$$

$$\Rightarrow x\in\left [ \dfrac {1}{3},\;1 \right]$$

### The domain of $$f(x) = cos^{-1}(3x-2)$$ is

A

$$x\in\left [ \dfrac {5}{4},\;7 \right]$$

.

B

$$x\in\left [ \dfrac {-2}{3},\;\dfrac {1}{3} \right]$$

C

$$x\in\left [ \dfrac {1}{3},\;1 \right]$$

D

$$x\in\left [ 2,\;\infty \right)$$

Option C is Correct

# Derivative of Inverse Cosine Function

• $$\dfrac {d}{dx}\,(cos^{-1}x)=\dfrac {-1}{\sqrt {1-x^2}}\;\;(-1\leq x \leq 1)$$

• $$\dfrac {d}{dx}\,cos^{-1}\Big(f(x)\Big)=\underbrace{\dfrac {-1}{\sqrt {1-(f(x))^2}}×f'(x)}_{chain\,rule}$$

$$=\dfrac {-f'(x)}{\sqrt {1-(f(x))^2}}$$

#### If $$y=cos^{-1}\Big(3x-1\Big)$$, find $$\dfrac {dy}{dx}$$.

A $$\dfrac {-3}{\sqrt {6x-9x^2}}$$

B $$\dfrac {-4}{\sqrt {4x-x^2}}$$

C $$2\,sin\,x+cos^{-1}\,x$$

D $$2\,cos\,x+sin^{-1}\,x$$

×

$$\dfrac {d}{dx}\,\Big(cos^{-1}f(x)\Big)=\dfrac {-f'(x)}{\sqrt {1-(f(x))^2}}$$

$$\therefore$$ $$\dfrac {d}{dx}\,\Big(cos^{-1}\,(3x-1)\Big) =\dfrac {-1}{\sqrt {1-(3x-1)^2}}×\dfrac {d}{dx}(3x-1)$$

$$\dfrac {d}{dx}\,\Big(cos^{-1}\,(3x-1)\Big) =\dfrac {-1}{\sqrt {1-(9x^2+1-6x)}}×3$$

$$\dfrac {d}{dx}\,\Big(cos^{-1}\,(3x-1)\Big)= \dfrac {-3}{\sqrt {6x-9x^2}}$$

### If $$y=cos^{-1}\Big(3x-1\Big)$$, find $$\dfrac {dy}{dx}$$.

A

$$\dfrac {-3}{\sqrt {6x-9x^2}}$$

.

B

$$\dfrac {-4}{\sqrt {4x-x^2}}$$

C

$$2\,sin\,x+cos^{-1}\,x$$

D

$$2\,cos\,x+sin^{-1}\,x$$

Option A is Correct