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# Differentiation of trigonometric functions for AQA A-level Maths

This page covers the following topics:

1. Differentiating trigonometric functions

2. Differentiating trigonometric reciprocals

3. Differentiating trigonometric functions with the chain rule

4. Differentiating trigonometric functions without the chain rule

Equations involving trigonometric functions can be differentiated using the basic rules of differentiation by using the following results: d(sinx)/dx = cosx, d(cosx)/dx = āsinx and d(tanx)/dx = secĀ²x.

Equations involving trigonometric reciprocals can be differentiated using the basic rules of differentiation by using the following results: d(cotx)/dx = ācosecĀ²x, d(secx)/dx = secxtanx and d(cosecx)/dx = ācosecxcotx.

The Chain rule can be applied to differentiate equations involving trigonometric functions using the results of their derivatives.

All differentiation rules can be used with the results for the derivatives of trigonometric functions to differentiate equations involving trigonometric functions.

# 1

Find a function for the slope for the given graph.

The equation of the graph is given by y = sinx/x. To find a function for the slope, this must be differentiated. This can be done using the Product rule, where u = 1/x = xā»Ā¹ and v = sinx. Then, du/dx = ā1/xĀ² and dv/dx = cosx. So, by the Product rule, dy/dx = āsinx/xĀ² + cosx/x.

# 2

Differentiate the following function: g(x) = secĀ²xcosx/tanx.

The function can be rewritten as g(x) = secĀ²xcosx/tanx = secxcotx. This can be differentiated using the Product rule. Let u = secx and v = cotx. Then, du/dx = secxtanx and dv/dx = ācosecĀ²x. So, by the Product rule, g'(x) = āsecxcosecĀ²x + cotxsecxtanx = āsecxcosecĀ²x + secx.

# 3

Prove that the derivative of cotx is ācosecĀ²x.

Let y = cotx = 1/tanx = (tanx)ā»Ā¹. So, using the Power rule, dy/dx = ā(tanx)ā»Ā²secĀ²x = āsecĀ²x/tanĀ²x = ācosĀ²x/cosĀ²xsinĀ²x = ā1/sinĀ²x = ācosecĀ²x.

# 4

Differentiate the following function: f(x) = 1/tanx + secx.

The function can be rewritten as f(x) = 1/tanx + secx = cotx + secx. Using results for the derivatives of trigonometric reciprocals, f'(x) = ācosecĀ²x + secxtanx.

# 5

Find the derivative of y = 8tanĀ³x.

This can be done by using the Chain rule twice. Let y = 8uĀ³ and u = tanx. Then, dy/du = 24uĀ² and du/dx = secĀ²x. So, dy/dx = 24uĀ²secĀ²x = 24tanĀ²xsecĀ²x.

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