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.

Illustration Questions

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

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

Illustration Questions

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

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 \(-1\leq g(x) \leq 1\).

Illustration Questions

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

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

 

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

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

Illustration Questions

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

Derivative of Inverse Tangent Function

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

\(=\dfrac {1}{1+(f(x))^2}×f'(x)\)

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

Illustration Questions

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 \(\left ( \dfrac {-\pi}{2}<y<\;\dfrac {\pi}{2} \right)\)

Illustration Questions

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

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