How to find the derivatives of inverse trigonometric functions?
We remark that inverse trigonometric functions are continuous functions. Now we use first principles and chain rule to find derivatives of these functions:
Derivative of f given by f(x) = sin–1 x.
From first principle
f(x) = sin–1 x and f(x+h) = sin–1 (x+h)
[Tex]\frac{d}{dx}(f(x)) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} \\ = \lim_{h\to0} \frac{sin^{-1}(x+h)-sin^{-1}x}{h}\\ = \lim_{h\to0} \frac{sin^{-1}[(x+h)\sqrt{1-x^2} – x\sqrt{1-(x+h)^2}]}{h}[/Tex]
Using the formula,
sin–1 x – sin–1 y = sin-1(x [Tex]\sqrt{1-y^2} [/Tex] – y [Tex]\sqrt{1-x^2} [/Tex])
[Tex]= \lim_{h\to0} \frac{sin^{-1}[(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}]}{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}} \times \frac{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}{h}[/Tex]
As
[Tex]\lim_{x\to0} (\frac{sin^{-1}x}{x}) = 1 \\ So, \\ = \lim_{h\to0} 1. \frac{(x+h)\sqrt{1-x^2}-x\sqrt{1-(x+h)^2}}{h} \\ = \lim_{h\to0} \frac{(x+h)^2(1-x^2)-x^2(1-(x+h)^2)}{h} \times \frac{1}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}\\ = \lim_{h\to0} \frac{(x+h)^2-x^2}{h} \times \frac{1}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}\\ = \lim_{h\to0} (2x+h) \times \frac{1}{(x+h)\sqrt{1-x^2}+x\sqrt{1-(x+h)^2}}\\ = \frac{(2x)}{2x \sqrt{1-x^2}}\\ = \frac{1}{\sqrt{1-x^2}} [/Tex]
Hence
[Tex]\frac{d}{dx} (sin^{-1}x) = \frac{1}{\sqrt{1-x^2}}[/Tex]
From chain rule
y = sin-1x
sin y = x
Differentiating both sides w.r.t x, we get
[Tex]\frac{d}{dx}(sin \hspace{0.1cm}y) = 1 \\ \frac{d}{dx}(sin \hspace{0.1cm}y) \times \frac{dy}{dx} = 1\\ \cos\hspace{0.1cm}y\frac{dy}{dx} = 1\\ \frac{dy}{dx} = \frac{1}{cos\hspace{0.1cm}y}\\ \frac{dy}{dx} = \frac{1}{\sqrt{1-sin^2\hspace{0.1cm}y}}[/Tex]
As
[Tex]\sin\hspace{0.1cm}y = x \\ \frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}}\\[/Tex]
Derivative of f given by f(x) = cos–1 x.
From first principle
f(x) = cos–1 x and f(x+h) = cos–1 (x+h)
[Tex]\frac{d}{dx}(f(x)) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} \\ = \lim_{h\to0} \frac{cos^{-1}(x+h)-cos^{-1}x}{h}\\ = \lim_{h\to0} \frac{(\frac{\pi}{2}-sin^{-1}(x+h))-(\frac{\pi}{2}-sin^{-1}x)}{h}[/Tex]
As
[Tex]cos^{-1} x = \frac{\pi}{2} – sin^{-1} x \\ = – \lim_{h\to0} \frac{sin^{-1}(x+h)-sin^{-1}x)}{h}[/Tex]
By using the previous result
[Tex]= – \frac{1}{\sqrt{1-x^2}}[/Tex]
Hence,
[Tex]\frac{d}{dx} (cos^{-1}x) = -\frac{1}{\sqrt{1-x^2}}[/Tex]
From chain rule
y = cos-1x
cos y = x
Differentiating both sides w.r.t x, we get
[Tex]\frac{d}{dx}(cos \hspace{0.1cm}y) = 1 \\ \frac{d}{dx}(cos \hspace{0.1cm}y) \times \frac{dy}{dx} = 1\\ -sin\hspace{0.1cm}y\frac{dy}{dx} = 1\\ \frac{dy}{dx} = – \frac{1}{sin\hspace{0.1cm}y}\\ \frac{dy}{dx} = -\frac{1}{\sqrt{1-cos^2\hspace{0.1cm}y}}[/Tex]
As
[Tex]cos\hspace{0.1cm}y = x \\ \frac{dy}{dx} = -\frac{1}{\sqrt{1-x^2}}\\[/Tex]
Derivative of f given by f(x) = tan–1 x.
From first principle
f(x) = tan–1 x and f(x+h) = tan–1 (x+h)
[Tex]\frac{d}{dx}(f(x)) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} \\ = \lim_{h\to0} \frac{tan^{-1}(x+h)-tan^{-1}x}{h}[/Tex]
As
[Tex]tan^{-1} x – tan^{-1}y = tan^{-1}(\frac{x-y}{1+xy}) \\ = \lim_{h\to0} \frac{tan^{-1}(\frac{x+h-x}{1+x(x+h)})}{h}\\ = \lim_{h\to0} \frac{tan^{-1}(\frac{h}{1+x^2+xh})}{\frac{h}{1+x^2+xh}} \times \frac{1}{1+x^2+xh}\\ = 1. \frac{1}{1+x^2}[/Tex]
Hence
[Tex]\frac{d}{dx} (tan^{-1}x)= \frac{1}{1+x^2}[/Tex]
From chain rule
y = tan-1x
tan y = x
Differentiating both sides w.r.t x, we get
[Tex]\frac{d}{dx}(tan \hspace{0.1cm}y) = 1 \\ \frac{d}{dx}(tan \hspace{0.1cm}y) \times \frac{dy}{dx} = 1\\ sec^2\hspace{0.1cm}y\frac{dy}{dx} = 1\\ \frac{dy}{dx} = \frac{1}{sec^2\hspace{0.1cm}y}\\ \frac{dy}{dx} = \frac{1}{1+tan^2\hspace{0.1cm}y}[/Tex]
As
tan y = x dy/dx = 1/1+x2
Derivative of f given by f(x) = cot–1 x.
From first principle
f(x) = cot–1 x and f(x+h) = cot–1 (x+h)
[Tex]\frac{d}{dx}(f(x)) = \lim_{h\to0} \frac{f(x+h)-f(x)}{h} \\ = \lim_{h\to0} \frac{cot^{-1}(x+h)-cot^{-1}x}{h}[/Tex]
As
[Tex]cot^{-1} x = \frac{\pi}{2} – tan^{-1}x\\ = \lim_{h\to0} \frac{(\frac{\pi}{2} – tan^{-1}(x+h))-(\frac{\pi}{2} – tan^{-1}x)}{h}\\ = – \lim_{h\to0} \frac{tan^{-1}(x+h)-tan^{-1}x}{h}[/Tex]
[Tex]= – \frac{1}{1+x^2}[/Tex]
Hence
[Tex] \frac{d}{dx} (cot^{-1}x)= \frac{-1}{1+x^2}[/Tex]
From chain rule
y = cot-1x
cot y = x
Differentiating both sides w.r.t x, we get
[Tex]\frac{d}{dx}(cot \hspace{0.1cm}y) = 1 \\ \frac{d}{dx}(cot \hspace{0.1cm}y) \times \frac{dy}{dx} = 1\\ -cosec^2\hspace{0.1cm}y\frac{dy}{dx} = 1\\ \frac{dy}{dx} = \frac{-1}{cosec^2\hspace{0.1cm}y}\\ \frac{dy}{dx} = \frac{-1}{1+cot^2\hspace{0.1cm}y}[/Tex]
As
cot y = x dy/dx = -1/1 + x2
Derivative of f given by f(x) = sec–1 x.
From chain rule
y = sec-1x
sec y = x
Differentiating both sides w.r.t x, we get
[Tex]\frac{d}{dx}(sec \hspace{0.1cm}y) = 1 \\ \frac{d}{dx}(sec \hspace{0.1cm}y) \times \frac{dy}{dx} = 1\\ sec\hspace{0.1cm}y \hspace{0.1cm}tan\hspace{0.1cm}y\frac{dy}{dx} = 1\\ \frac{dy}{dx} = \frac{1}{sec\hspace{0.1cm}y \hspace{0.1cm}tan\hspace{0.1cm}y}\\ \frac{dy}{dx} = \frac{1}{|sec\hspace{0.1cm}y| \hspace{0.1cm}|tan\hspace{0.1cm}y|}\\[/Tex]
If x > 1, then y ∈ (0, π/2)
∴ sec y > 0, tan y > 0 ⇒ |sec y||tan y| = sec y tan y
If x < -1, then y ∈ (π/2,π)
∴ sec y < 0, tan y < 0 ⇒ |sec y||tan y| = (-sec y) (-tan y) = sec y tan y
[Tex]\frac{dy}{dx} = \frac{1}{|sec \hspace{0.1cm}y| \sqrt{tan^2y}}\\ \frac{dy}{dx} = \frac{1}{|sec \hspace{0.1cm}y| \sqrt{1-sec^2y}}[/Tex]
As
sec y = x
[Tex]\frac{dy}{dx} = \frac{1}{|x|\sqrt{x^2-1}}\\[/Tex]
Derivative of f given by f(x) = cosec–1 x.
From chain rule
y = cosec-1x
cosec y = x
Differentiating both sides w.r.t x, we get
[Tex]\frac{d}{dx}(cosec \hspace{0.1cm}y) = 1 \\ \frac{d}{dx}(cosec \hspace{0.1cm}y) \times \frac{dy}{dx} = 1\\ -cosec\hspace{0.1cm}y \hspace{0.1cm}cot\hspace{0.1cm}y\frac{dy}{dx} = 1\\ \frac{dy}{dx} = \frac{1}{-cosec\hspace{0.1cm}y \hspace{0.1cm}cot\hspace{0.1cm}y}\\ \frac{dy}{dx} = \frac{-1}{|cosec\hspace{0.1cm}y| \hspace{0.1cm}|cot\hspace{0.1cm}y|}\\[/Tex]
If x > 1, then y ∈ (0, π/2)
∴ cosec y > 0, cot y > 0 ⇒ |cosec y||cot y| = cosec y cot y
If x < -1, then y∈ (-π/2, 0)
∴ cosec y < 0, cot y < 0 ⇒ |cosec y||cot y| = (-cosec y) (-cot y)
[Tex]\frac{dy}{dx} = \frac{-1}{|cosec \hspace{0.1cm}y| \sqrt{cot^2y}}\\ \frac{dy}{dx} = \frac{-1}{|cosec \hspace{0.1cm}y| \sqrt{cosec^2y-1}}[/Tex]
As
cosec y = x
[Tex]\frac{dy}{dx} = \frac{-1}{|x|\sqrt{x^2-1}}\\[/Tex]
Also, Check
Derivatives of Inverse Functions
In mathematics, a function(e.g. f), is said to be an inverse of another(e.g. g), if given the output of g returns the input value given to f. Additionally, this must hold true for every element in the domain co-domain(range) of g. E.g. assuming x and y are constants if g(x) = y and f(y) = x then the function f is said to be an inverse of the function g. Or in other words, if a function f : A ⇢ B is one – one and onto function or bijective function, then a function defined by g : B ⇢ A is known as inverse of function f. The inverse function is also known as the anti function. The inverse of function is denoted by f-1.
f(g(x)) = g(f(x)) = x
Here, f and g are inverse functions.
Table of Content
- Overview of Derivatives of Inverse Functions
- Procedure of finding inverse of f
- Derivatives of Inverse Functions
- How to find derivatives of inverse functions from the table?
- Derivatives of Inverse Trigonometric Functions
- How to find the derivatives of inverse trigonometric functions?
- 1. Derivative of f given by f(x) = sin–1 x.
- 2. Derivative of f given by f(x) = cos–1 x.
- 3. Derivative of f given by f(x) = tan–1 x.
- 4. Derivative of f given by f(x) = cot–1 x.
- 5. Derivative of f given by f(x) = sec–1 x.
- 6. Derivative of f given by f(x) = cosec–1 x.
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