7.3.6 Differentiating Reciprocal and Inverse Trig Functions

Differentiating reciprocal and inverse trigonometric functions is an important topic in calculus, particularly when dealing with more complex problems involving trigonometric functions. Here's a detailed explanation of how to differentiate these functions.

1. Differentiating Reciprocal Trigonometric Functions:

Reciprocal trigonometric functions are the reciprocals of the basic trigonometric functions: sine, cosine, and tangent. The main reciprocal functions are cosecant $\csc(x)$ , secant $\sec(x)$ , and cotangent $\cot(x).$

Reciprocal Function Derivatives:

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Cosecant $\csc(x)$ : $\frac{d}{dx} \csc(x) = \frac{d}{dx} \left(\frac{1}{\sin(x)}\right) = -\csc(x) \cot(x)$ Derivation using the quotient rule:

Let $\ u(x) = 1 \ and \ v(x) = \sin(x) .$
Apply the quotient rule: $\frac{d}{dx} \csc(x) = \frac{0 \cdot \sin(x) - 1 \cdot \cos(x)}{\sin^2(x)} = -\frac{\cos(x)}{\sin^2(x)} = -\csc(x) \cot(x)$

Secant $\sec(x)$ : $\frac{d}{dx} \sec(x) = \frac{d}{dx} \left(\frac{1}{\cos(x)}\right) = \sec(x) \tan(x)$ Derivation using the quotient rule:

Let $\ u(x) = 1 \ and \ v(x) = \cos(x) .$
Apply the quotient rule: $\frac{d}{dx} \sec(x) = \frac{0 \cdot \cos(x) + 1 \cdot \sin(x)}{\cos^2(x)} = \frac{\sin(x)}{\cos^2(x)} = \sec(x) \tan(x)$

Cotangent $\cot(x)$ : $\frac{d}{dx} \cot(x) = \frac{d}{dx} \left(\frac{1}{\tan(x)}\right) = -\csc^2(x)$ Derivation using the quotient rule:

Let $\ u(x) = 1 \ and \ v(x) = \tan(x) .$
Apply the quotient rule: $\frac{d}{dx} \cot(x) = \frac{0 \cdot \tan(x) - 1 \cdot \sec^2(x)}{\tan^2(x)} = -\frac{\sec^2(x)}{\tan^2(x)} = -\csc^2(x)$

2. Differentiating Inverse Trigonometric Functions:

Inverse trigonometric functions are the inverses of the basic trigonometric functions: sine, cosine, and tangent. The main inverse functions are arcsine $\arcsin(x)$ , arccosine $\arccos(x)$ , and arctangent $\arctan(x).$

Inverse Function Derivatives:

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Arcsine $\arcsin(x)$ : $\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}}$

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Arccosine $\ arccos(x):$ $\frac{d}{dx} \arccos(x) = -\frac{1}{\sqrt{1 - x^2}}$

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Arctangent $\arctan(x)$ : $\frac{d}{dx} \arctan(x) = \frac{1}{1 + x^2}$

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Arccotangent $\ arccot(x)$ : $\frac{d}{dx} \ arccot(x) = -\frac{1}{1 + x^2}$

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Arcsecant $\ arcsec(x):$ $\frac{d}{dx} \ arcsec(x) = \frac{1}{|x| \sqrt{x^2 - 1}}$

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Arccosecant $\ arccsc(x)$ : $\frac{d}{dx} \ arccsc(x) = -\frac{1}{|x| \sqrt{x^2 - 1}}$

3. Summary of Derivatives:

Here is a summary of the key derivatives:

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Reciprocal Trig Functions:

$\frac{d}{dx} \csc(x) = -\csc(x) \cot(x)$

$\frac{d}{dx} \sec(x) = \sec(x) \tan(x)$

$\frac{d}{dx} \cot(x) = -\csc^2(x)$

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Inverse Trig Functions:

$\frac{d}{dx} \arcsin(x) = \frac{1}{\sqrt{1 - x^2}}$

$\frac{d}{dx} \arccos(x) = -\frac{1}{\sqrt{1 - x^2}}$

$\frac{d}{dx} \arctan(x) = \frac{1}{1 + x^2}$

$\frac{d}{dx} \ arccot(x) = -\frac{1}{1 + x^2}$

$\frac{d}{dx} \ arcsec(x) = \frac{1}{|x| \sqrt{x^2 - 1}}$

$\frac{d}{dx} \ arccsc(x) = -\frac{1}{|x| \sqrt{x^2 - 1}}$

Differentiating Reciprocal and Inverse Trig Functions Simplified Revision Notes for A-Level OCR Maths Pure

7.3.6 Differentiating Reciprocal and Inverse Trig Functions

1. Differentiating Reciprocal Trigonometric Functions:

Reciprocal Function Derivatives:

2. Differentiating Inverse Trigonometric Functions:

Inverse Function Derivatives:

3. Summary of Derivatives:

Reciprocal Trig Functions:

Inverse Trig Functions:

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