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Derivative of arcsin x is 1/√1-x². It is written as d/dx(arcsin x) = 1/√1-x². Arcsin function is the inverse of the sine function and is a pure trigonometric function. We will learn how to differentiate arcsin x by using various differentiation rules like the first principle of derivative, differentiate arcsin x using chain rule and differentiate arcsin x using the quotient rule. Arcsin of x is defined as the inverse sine function of x when -1 ≤ x ≤ 1.

Derivative of arcsin function is denoted by d/dx(arcsin x) and its value is 1/√1-x². It returns the angle whose sine is a given number.

When the sine of y is equal to x:

sin y = x

Then the arcsine of x is equal to the inverse sine function of x, which is equal to y:

\(arcsin x = sin^{-1}x = y\)

Example: \(arcsin 1 = sin^{-1} 1 = π/2 rad = 90°\)

Graph of Arcsine: Arcsin x can be represented in graphical form as follows:

Values of Arcsin

x | arcsin(x) (rad) | arcsin(x) (°) |

-1 | -π/2 | -90° |

\(-\sqrt3\over2\) | -π/3 | -60° |

\(-\sqrt2\over2\) | -π/4 | -45° |

\(-1\over2\) | -π/6 | -30° |

0 | 0 | 0° |

\(1\over2\) | π/6 | 30° |

\(\sqrt2\over2\) | π/4 | 45° |

\(\sqrt3\over2\) | π/3 | 60° |

1 | π/2 | 90° |

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We will learn how to differentiate arcsin x by using various differentiation rules:

We can prove the derivative of arcsin by quotient rule using the following steps:

Step 1: Write sin y = x,

Step 2: Differentiate both sides of this equation with respect to x.

\(\begin{matrix}

{d\over{dx}}sin y = {d\over{dx}}x\\

cosy {d\over{dx}} y = 1

\end{matrix}\)

Step 3: Solve for \({dy\over{dx}}\)

\({d\over{dx}} y = {1\over{cosy}}\)

Step 4: Define cos y in terms of x using a reference triangle.

From the reference triangle, the adjacent side is \(\sqrt{(1 – x^2)}\) and the hypotenuse is 1. Thus, \(cos y = {\sqrt{(1 – x^2)}\over{1}}\) which means \({1\over{cosy}} = {1\over{\sqrt{(1 – x^2)}}}\)

Step 5: Substitute for cos y.

\({d\over{dx}} y = {1\over{cosy}} = {1\over{\sqrt{(1 – x^2)}}}\)

Step 6: Define arcsine.

Now we can define arcsine as:

\(y = sin^{-1}x\)

Step 7: Differentiate and write the result.

\({d\over{dx}}sin^{-1}x = {dy\over{dx}} = {1\over{\sqrt{(1 – x^2)}}}\)

We can prove the derivative of arcsin by Chain rule using the following steps:

\(\begin{matrix}

\text{ Let } y = arcsin x\\

\text{ Taking sin on both sides, }\\

sin y = sin (arcsin x)\\

\text{ By the definition of an inverse function, we have, }\\

sin (arcsin x) = x\\

\text{ So the equation becomes }

sin y = x \\

\text{ Differentiating both sides with respect to x,}\\

{d\over{dx}} (sin y) = {d\over{dx}} (x)\\

cosy {d\over{dx}} y = 1\\

{d\over{dx}} y = {1\over{cosy}}\\

\text{ Using one of the trigonometric identities }\\

sin^y + cos^y = 1\\

{\therefore} cos y = \sqrt{1 – sin^2y} = \sqrt{1 – x^2}\\

{dy\over{dx}} = {1\over{\sqrt{(1 – x^2)}}}\\

\text{ Substituting y = arcsin x back }\\

{d\over{dx}}(arcsin x) = arcsin’x = {1\over{\sqrt{1-x^2}}}

\end{matrix}\)

We can prove the derivative of arcsin by First Principle using the following steps:

\(\begin{matrix}

f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}

f(x)=arcsin x\\

f(x+h)=arcsin(x+h)\\

f(x+h)–f(x)= tan(x+h) – tan(x) = arcsin (x + h) – arcsin x\\

{f(x+h) – f(x)\over{h}}={ arcsin (x + h) – arcsin x\over{h}}\\

\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{h{\rightarrow}0} {arcsin (x + h) – arcsin x\over{h}}\\

\text{ Assume that arcsin (x + h) = A and arcsin x = B }\\

sin A = x + h\\

sin B = x\\

sin A – sin B = (x + h) – x\\

sin A – sin B = h\\

If \text{ h → 0, (sin A – sin B) → 0 sin A → sin B or A → B or A – B → 0}\\

\lim _{A-B{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{A-B{\rightarrow}0} {(A – B)\over{(sin A – sin B)}}\\

\text{ sin A – sin B = 2 sin [(A – B)/2] cos [(A + B)/2] }\\

\lim _{A-B{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{A-B{\rightarrow}0} {(A – B)\over{[2 sin [(A – B)/2] cos [(A + B)/2]]}}\\

\text{ A – B → 0, we can have (A – B)/2 → 0 }\\

\lim _{{A-B\over{2}}{\rightarrow}0}{f(x+h) –f(x)\over{h}} = \lim _{{A-B\over{2}}{\rightarrow}0}{1\over{(sin [(A – B)/2]\over{[(A – B)/2])}}} \lim _{A-B{\rightarrow}0} cos[(A + B)/2]\\

f’(x) = cos[(B + B)/2] = cos B\\

sin B = x\\

cos B = \sqrt{1 – sin^2B} = \sqrt{1 – x²}\\

\lim _{h{\rightarrow}0}{f(x+h) –f(x)\over{h}} = {1\over{\sqrt{(1 – x^2)}}}\\

f’(x)={dy\over{dx}} = {1\over{\sqrt{(1 – x^2)}}}

\end{matrix}\)

Rule name | Rule |

Sine of arcsine | sin( arcsin x ) = x |

Arcsine of sine | arcsin( sin x ) = x+2kπ, when k∈ℤ (k is integer) |

Arcsin of negative argument | arcsin(-x) = – arcsin x |

Complementary angles | arcsin x = π/2 – arccos x = 90° – arccos x |

Arcsin sum | \(arcsin\alpha + arcsin\beta = arcsin(\alpha\sqrt{(1-\beta^2)} + \beta\sqrt{(1-\alpha^2)})\) |

Arcsin difference | \(arcsin\alpha – arcsin\beta = arcsin(\alpha\sqrt{(1-\beta^2)} – \beta\sqrt{(1-\alpha^2)})\) |

Cosine of arcsine | \(cos(arcsin x) = sin(arccosx) = \sqrt{1-x^2}\) |

Tangent of arcsine | \(tan(arcsin x) = {x\over{\sqrt{1-x^2}}}\) |

Derivative of arcsine | \({d\over{dx}}(arcsin x) = arcsin’x = {1\over{\sqrt{1-x^2}}}\) |

Indefinite integral of arcsine | \(\int(arcsin x) dx = x(arcsin x) + {\sqrt{1-x^2}} + C\) |

Corollary | \(\dfrac {d {\arcsin {\frac x a} } } {d x} = \dfrac 1 {\sqrt {a^2 – x^2} }\) |

Example 1: What is the derivative of arcsin(x − 1)?

Solution: Derivative of inverse trigonometric functions. The general formula to differentiate the arcsin functions is

\(\begin{matrix}

\int{sin^{-1}u} = {1\over{\sqrt{(1 – u^2)}}} {du\over{dx}}\\

{d\over{dx}} sin^{-1}(x-1)={1\over{\sqrt{(1-(x-1)^2)}}}\times{d(x-1)\over{dx}}\\

{d\over{dx}} sin^{-1}(x-1)={1\over{\sqrt{(1-(x-1)^2)}}}

\end{matrix}\)

Example 2: What is the derivative of arcsin(x\a)?

Solution:

To start off, let’s set this function equal to y

\(\begin{matrix}

y=sin^{−1}({x\over{a}})\\

siny=({x\over{a}})\\

\text{ Multiply a to both sides and taking the derivative }\\

{d\over{dx}}[asiny] = {d\over{dx}}\\

{dy\over{dx}} acosy = 1\\

{dy\over{dx}} = {1\over{acosy}}\\

\text{ Divide both sides to isolate }{dy\over{dx}}\\

{dy\over{dx}} = {1\over{a}}secy\\

secy = {1\over{cosy}}\\

\text{ from the image below}\\

secy = {a\over{\sqrt{a2^−x^2}}}\\

\text{ We will now substitute this value back into the answer for our derivative: }

{dy\over{dx}} = {1\over{a}}secy\\

{dy\over{dx}} = {1\over{a}}{a\over{\sqrt{a2^−x^2}}}\\

\text{ Canceling out the a in the numerator and denominator, we are left with our final answer: }\\

{dy\over{dx}} = {1\over{\sqrt{a2^−x^2}}}

\end{matrix}\)

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