Further Differentiation Revision Notes for AQA A-Level Mathematics

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:

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

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

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

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

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

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

Further Differentiation (AQA A-Level Mathematics): Revision Notes

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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