Question

derivative of sin (1/x)​

Answer 1

$\sf \: Answer$

The derivative of sin inverse x is 1/√(1-x²) [where -1 < x < 1]

More to know :eyes:

Now, we will write the derivative of sin inverse x mathematically. The derivative of a function is the slope of the tangent to the function at the point of contact. Hence, 1/√(1-x²) is the slope function of the tangent to the graph of sin inverse x at the point of contact. Mostly, we memorize the derivative of sin inverse x. An easy way to do that is knowing the fact that the derivative of sin inverse x is the negative of the derivative of cos inverse x and the derivative of cos inverse x is the negative of the derivative of sin inverse x. The mathematical expression to write the differentiation of sin-1x is:

d(sin 1/x )/ dx = 1/√(1-x²) (-1 < x < 1)

Proof of Derivative of Sin Inverse x

We know that the derivative of sin inverse x is 1/√(1-x²), where -1 < x < 1. Now, we will derive the derivative using some differentiation formulas. To derive the derivative of sin inverse x, we will use the following formulas:

Now it must be well known to you

cos²θ + sin²θ = 1

Chain rule: (f(g(x)))' = f'(g(x)).g'(x)

d(sin x)/dx = cos x

Assume y = sin-1x ⇒ sin y = x

Differentiating both sides of sin y = x w.r.t. x, we have

cos y dy/dx = 1

⇒ dydx = 1/cos y

⇒ dy/dx = 1/√(1 - sin²y) (Using cos²θ + sin²θ = 1)

⇒ dy/dx = 1/√(1 - x²) (Because sin y = x)

⇒ d(sin-1/x)/dx = 1/√(1 - x²)

Hence, d(sin-1/x)/dx = 1/√(1 - x²) , that is, derivative of sin inverse x is 1/√(1 - x²)