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Single Variable Calculus
Differentiation
01The Derivative as the Slope of the Tangent Line02Secant Lines and the Limit Definition of the Derivative03Differentiating 1/x Straight From the Definition04Tangents to 1/x and the Triangle of Area 205Newton and Leibniz Notation, and the Power RuleProblem set0/10Problem set 20/10MIT problem set0/3Practice∞
01The Derivative as an Instantaneous Rate of Change02Rates of Change Without Time: Gradient and Sensitivity03Easy Limits, 0/0, and One-Sided Limits04Continuity, Jumps, and Removable Discontinuities05Infinite Discontinuities: 1/x and Its Derivative06Differentiable Implies ContinuousProblem set0/10Problem set 20/10Practice∞
01Two Kinds of Formula, and the Derivative of Sine02The Derivative of Cosine, and Two Limits at Zero03Bow and Bowstring: Proving the Two Trig Limits04Why the Trig Derivatives Hold Only in Radians05A Geometric Proof: Sine as a Height on the Circle06Finishing the Proof, and the Product and Quotient RulesProblem set0/10Problem set 20/10MIT problem set0/1Practice∞
01Proving the Product Rule: Change One Factor at a Time02The Quotient Rule, and the Power Rule for Negative Exponents03The Chain Rule: Differentiating a Function of a Function04Higher Derivatives: The Sine Cycle and Three Notations05The nth Derivative of xⁿ Is n Factorial, by InductionProblem set0/10Problem set 20/10MIT problem set0/2Practice∞
01The Power Rule for Rational Exponents02The Slope of a Circle, Two Ways03Implicit Differentiation of a Quartic Curve04Inverse Functions and the Reflection Across y = x05Derivatives of the Arctangent and ArcsineProblem set0/10Problem set 20/10MIT problem set0/1Practice∞

Derivatives of the Arctangent and Arcsine

How does one right triangle turn cos^2(arctan x) into 1/(1 + x^2), and what makes the square root in the arcsine derivative positive?


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    1. 1

      The derivative of tan⁡y\tan ytany and implicit differentiation

      Derive ddytan⁡y=1cos⁡2y=sec⁡2y\frac{d}{dy}\tan y = \frac{1}{\cos^2 y} = \sec^2 ydyd​tany=cos2y1​=sec2y with the quotient rule, then differentiate tan⁡y=x\tan y = xtany=x with respect to xxx and solve for y′y'y′.

    2. 2

      The right triangle that simplifies cos⁡2(arctan⁡x)\cos^2(\arctan x)cos2(arctanx)

      Draw the triangle with side xxx opposite yyy, adjacent side 111 and hypotenuse 1+x2\sqrt{1 + x^2}1+x2​, read off cos⁡y\cos ycosy, and finish with ddxarctan⁡x=11+x2\frac{d}{dx}\arctan x = \frac{1}{1 + x^2}dxd​arctanx=1+x21​.

    3. 3

      The derivative of arcsin⁡x\arcsin xarcsinx and its branch

      Differentiate sin⁡y=x\sin y = xsiny=x implicitly, rewrite cos⁡y\cos ycosy as 1−x2\sqrt{1 - x^2}1−x2​, state ddxarcsin⁡x=11−x2\frac{d}{dx}\arcsin x = \frac{1}{\sqrt{1 - x^2}}dxd​arcsinx=1−x2​1​, and record the branch choice behind the positive square root.

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