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Last updated on October 11, 2025
We explore the derivative of arc functions such as arcsin, arccos, and arctan, which are crucial in understanding how the inverse trigonometric functions change concerning slight changes in x. These derivatives are widely used in various real-world applications, from calculating angles in physics to solving integrals in mathematics. We will now delve into the derivatives of inverse trig functions in detail.
The derivative of arc functions, namely arcsin, arccos, and arctan, are essential in calculus. They are represented as d/dx (arcsin x), d/dx (arccos x), and d/dx (arctan x), respectively. Each function has a well-defined derivative, indicating that they are differentiable within their domains.
The key concepts are mentioned below:
Arcsin Function: The inverse of the sine function.
Arccos Function: The inverse of the cosine function.
Arctan Function: The inverse of the tangent function.
The derivatives of arc functions are represented as follows: d/dx (arcsin x) = 1/√(1-x²) d/dx (arccos x) = -1/√(1-x²) d/dx (arctan x) = 1/(1+x²)
These formulas apply to their respective domains, where x is within the interval that ensures the expressions under the square root are non-negative.
We can derive the derivatives of arc functions using proofs. To show this, we utilize trigonometric identities along with the rules of differentiation. Here are the methods used to prove the derivatives of arcsin, arccos, and arctan:
We will demonstrate that the differentiation of these arc functions results in their respective derivatives using implicit differentiation: Derivative of Arcsin x Let y = arcsin x, then sin y = x. Differentiating both sides with respect to x, we get: cos y · dy/dx = 1 dy/dx = 1/cos y Using the identity cos²y = 1 - sin²y and sin y = x, we have cos y = √(1 - x²). Thus, dy/dx = 1/√(1 - x²).
Derivative of Arccos x Let y = arccos x, then cos y = x. Differentiating both sides with respect to x, we get: -sin y · dy/dx = 1 dy/dx = -1/sin y Using the identity sin²y = 1 - cos²y and cos y = x, we have sin y = √(1 - x²). Thus, dy/dx = -1/√(1 - x²). Derivative of Arctan x Let y = arctan x, then tan y = x. Differentiating both sides with respect to x, we get: sec²y · dy/dx = 1 dy/dx = 1/sec²y Using the identity sec²y = 1 + tan²y and tan y = x, we have sec²y = 1 + x². Thus, dy/dx = 1/(1 + x²).
When a function is differentiated multiple times, the resulting derivatives are known as higher-order derivatives. Higher-order derivatives of arc functions provide insights akin to analyzing the acceleration (second derivative) and jerk (third derivative) in motion.
For the first derivative of a function, we write f′(x), indicating how the function changes or its slope at a certain point. The second derivative, f′′(x), results from differentiating the first derivative and continues in this pattern for higher-order derivatives.
Certain values of x may cause the derivative of arc functions to be undefined: For arcsin x and arccos x, the derivatives are undefined at x = ±1 because the square root in the denominator becomes zero.
For arctan x, there are no points where the derivative is undefined within its domain.
Students often encounter errors when differentiating arc functions. Understanding the correct methods can resolve these mistakes. Here are common mistakes and how to avoid them:
Calculate the derivative of (arcsin x · x²)
Here, we have f(x) = arcsin x · x². Using the product rule, f'(x) = u′v + uv′ In the given equation, u = arcsin x and v = x². Let’s differentiate each term, u′ = d/dx (arcsin x) = 1/√(1-x²) v′ = d/dx (x²) = 2x Substituting into the given equation, f'(x) = (1/√(1-x²)) · x² + arcsin x · 2x Let’s simplify terms to get the final answer, f'(x) = x²/√(1-x²) + 2x arcsin x Thus, the derivative of the specified function is x²/√(1-x²) + 2x arcsin x.
We find the derivative of the given function by dividing it into two parts.
The first step involves finding the derivative of each part, then combining them using the product rule to get the final result.
A satellite's trajectory is modeled by the function y = arctan(x), where y represents the angle from the ground at a distance x. If x = 1, determine the rate of change of the angle.
We have y = arctan(x) (angle of the satellite)...(1) Now, we will differentiate equation (1) Take the derivative of arctan(x): dy/dx = 1/(1+x²) Given x = 1, substitute this into the derivative dy/dx = 1/(1+1²) dy/dx = 1/2 Hence, the rate of change of the angle at x = 1 is 1/2.
We calculate the rate of change of the angle at x = 1, indicating how the angle varies with respect to changes in distance.
Derive the second derivative of the function y = arcsin(x).
The first step is to find the first derivative, dy/dx = 1/√(1-x²)...(1) Now we will differentiate equation (1) to get the second derivative: d²y/dx² = d/dx [1/√(1-x²)] Using the chain rule, d²y/dx² = -(1-x²)^(-3/2) · (-2x) d²y/dx² = 2x/(1-x²)^(3/2) Therefore, the second derivative of the function y = arcsin(x) is 2x/(1-x²)^(3/2).
We utilize the chain rule, where the first derivative is differentiated to find the second derivative.
We simplify using the chain rule and trigonometric identities to find the final answer.
Prove: d/dx (arctan²(x)) = 2 arctan(x)/(1+x²).
Let’s start using the chain rule: Consider y = arctan²(x) [arctan(x)]² To differentiate, we use the chain rule: dy/dx = 2 arctan(x) · d/dx [arctan(x)] Since the derivative of arctan(x) is 1/(1+x²), dy/dx = 2 arctan(x) · 1/(1+x²) Substituting y = arctan²(x), d/dx (arctan²(x)) = 2 arctan(x)/(1+x²) Hence proved.
In this step-by-step process, we use the chain rule to differentiate the equation.
Then, we replace arctan(x) with its derivative.
Finally, we substitute y = arctan²(x) to derive the equation.
Solve: d/dx (arcsin(x)/x)
To differentiate the function, we use the quotient rule: d/dx (arcsin(x)/x) = (d/dx (arcsin(x)) · x - arcsin(x) · d/dx(x))/x² Substitute d/dx (arcsin(x)) = 1/√(1-x²) and d/dx(x) = 1 (1/√(1-x²) · x - arcsin(x) · 1)/x² = (x/√(1-x²) - arcsin(x))/x² Therefore, d/dx (arcsin(x)/x) = (x/√(1-x²) - arcsin(x))/x²
In this process, we differentiate the given function using the quotient rule.
As a final step, we simplify the equation to obtain the final result.
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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