Derivative of Arccos

The derivative of arccos is the differentiation of the inverse cosine function arccos x which is -1/√(1-x²) where -1 < x < 1. It is written as the derivative of arccos x or derivative of cos inverse x, denoted by, d(arccos x)/dx = d(cos^-1x)/dx = -1/√(1-x²). The derivative of arccos is the process of determining the rate of change of arccos x with respect to the variable x. Hence, the rate of change of arccos x at a particular angle, that is, the derivative of arccos x is given by -1/√(1-x²).

In this article, we will derive the derivative of arccos using different methods including implicit differentiation and the first principle of differentiation. We also discuss the anti-derivative of arccos x along with some examples using the derivative of arccos.

The derivative of arccos x is given by -1/√(1-x²) where -1 < x < 1. It is also called the derivative of cos inverse x, that is, the derivative of the inverse cosine function. Derivatives of all inverse trigonometric functions can be calculated using the method of implicit differentiation. Since the derivative of arccos x is -1/√(1-x²), therefore the graph of the derivative of cos inverse x will be the graph of -1/√(1-x²).

Derivative of Arccos Formula

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 arccos x at the point of contact. An easy way to do memorize the derivative of arccos x is knowing the fact that the derivative of arccos x is the negative of the derivative of sin inverse x and the derivative of sin inverse x is the negative of the derivative of arccos x. Now, we will write the derivative of arccos x mathematically. The mathematical expression to write the differentiation of cos^-1x is:

d(arccos x)/dx = d(cos^-1x )/ dx = -1/√(1-x²), where -1 < x < 1

Now, we will prove the derivative of arcos x using some trigonometric formulas and identities. Assume y = cos^-1x ⇒ cos y = x. Differentiate both sides of the equation cos y = x with respect to x using the chain rule.

cos y = x

⇒ d(cos y)/dx = dx/dx

⇒ -sin y dy/dx = 1

⇒ dy/dx = -1/sin y ---- (1)

Since cos²y + sin²y = 1, we have sin y = √(1 - cos²y) = √(1 - x²) [Because cos y = x]

Substituting sin y = √(1 - x²) in (1), we have

dy/dx = -1/√(1 - x²)

Since x = -1, 1 makes the denominator √(1 - x²) equal to 0, and hence the derivative is not defined at x = -1 and x = 1, therefore x cannot be -1 and 1.

Hence the derivative of arccos x is -1/√(1 - x²), where -1 < x < 1

Now, we will prove the derivative of arccos using the first principle of differentiation. To prove, we will use some differentiation formulas, inverse trigonometric formulas, and identities such as:

• \(f'(x)=\lim_{h\rightarrow 0}\dfrac{f(x+h)-f(x)}{h}\)
• arccos x + arcsin x = π/2 ⇒ arccos x = π/2 - arcsin x
• \(\frac{\mathrm{d} \arcsin x}{\mathrm{d} x}=\lim_{h\rightarrow 0}\dfrac{\arcsin(x+h)-\arcsin x}{h}\)

Let us begin the proof of the derivative of arccos x using the first principle of differentiation.

\(\begin{align} \frac{\mathrm{d} \arccos x}{\mathrm{d} x}&=\lim_{h\rightarrow 0}\dfrac{\arccos(x+h)-\arccos x}{h}\\&=\lim_{h\rightarrow 0}\dfrac{\frac{\pi}{2}-\arcsin(x+h)-(\frac{\pi}{2}-\arcsin x)}{h}\\&= \lim_{h\rightarrow 0}\dfrac{\frac{\pi}{2}-\arcsin(x+h)-\frac{\pi}{2}+\arcsin x}{h}\\&=\lim_{h\rightarrow 0}\dfrac{-\arcsin(x+h)+\arcsin x}{h} \\&=\lim_{h\rightarrow 0}-\dfrac{(\arcsin(x+h)-\arcsin x)}{h}\\&=-\lim_{h\rightarrow 0}\dfrac{\arcsin(x+h)-\arcsin x}{h}\\&=-\dfrac{1}{\sqrt{1-x^2}}\end{align}\)

Hence, we have proved the derivative of arccos using the definition of limits, that is, the first principle of differentiation.

Now, that we have derived the derivative of arccos, we will find the anti-derivative of arccos, that is, ∫arccos x dx = ∫cos^-1 x dx using the integration by parts (ILATE).

∫cos^-1x = ∫cos^-1x · 1 dx

Using integration by parts,

∫f(x) . g(x) dx = f(x) ∫g(x) dx − ∫(f′(x) ∫g(x) dx) dx + C

Here f(x) = cos^-1x and g(x) = 1.

∫cos^-1x · 1 dx = cos^-1x ∫1 dx - ∫ [d(cos^-1x)/dx ∫1 dx]dx + C

∫cos^-1x dx = cos^-1x . (x) - ∫ [-1/√(1 - x²)] x dx + C

We will evaluate this integral ∫ [-1/√(1 - x²)] x dx using substitution method. Assume 1-x² = u. Then -2x dx = du (or) x dx = -1/2 du.

∫cos^-1x dx = x cos^-1x - ∫(-1/√u) (-1/2) du + C

= x cos^-1x - 1/2 ∫u^-1/2 du + C

= x cos^-1x - (1/2) (u^1/2/(1/2)) + C

= x cos^-1x - √u + C

= x cos^-1x - √(1 - x²) + C

Therefore, ∫cos^-1x dx = x cos^-1x - √(1 - x²) + C

We know that the derivative of arccos x, that is, cos inverse x is -1/√(1 - x²) and the derivative of sin inverse x is 1/√(1 - x²). To determine the derivative of cos inverse w.r.t. sin inverse, we will divide the derivatives of both the functions.

d(cos^-1x)/dx = -1/√(1 - x²) --- (1)

d(sin^-1x)/dx = 1/√(1 - x²) --- (2)

Divide (1) and (2), we have

[d(cos^-1x)/dx]/[d(sin^-1x)/dx] = [-1/√(1 - x²)]/[1/√(1 - x²)]

⇒ d(cos^-1x)/d(sin^-1x) = -1

Hence, the derivative of cos inverse w.r.t. sin inverse is -1.

Important Notes on Derivative of Arccos

• The derivative of arccos x is given by -1/√(1-x²) where -1 < x < 1
• The derivative of cos inverse w.r.t. sin inverse is -1.
• ∫cos^-1x dx = x cos^-1x - √(1 - x²) + C

Topics Related to Derivative of Arccos

• Derivative of Sin inverse x
• Cos Inverse Formula
• Inverse Trigonometric Formulas

What is Derivative of Arccos in Trigonometry?

The derivative of arccos x in trigonometry is given by -1/√(1-x²) where -1 < x < 1.

How to Prove the Derivative of Arccos x?

The derivative of arccos can be proved using different methods including implicit differentiation and the first principle of differentiation.

What is the Anti-Derivative of Arccos?

The anti-derivative of arccos is nothing but the integral of the inverse cosine function which is given by ∫cos^-1x dx = x cos^-1x - √(1 - x²) + C

What is the Derivative of Arccos w.r.t. Sin Inverse?

The derivative of cos inverse w.r.t. sin inverse is -1.

How to Find the Derivative of Arccos x?

We can obtain the derivative of arccos using different methods including implicit differentiation and the first principle of differentiation.