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Differentiation of trigonometric functions for Edexcel 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 = sec²x.

Equations involving trigonometric reciprocals can be differentiated using the basic rules of differentiation by using the following results: d(cotx)/dx = ācosec²x, 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/x² and dv/dx = cosx. So, by the Product rule, dy/dx = āsinx/x² + cosx/x.

2
Differentiate the following function: g(x) = sec²xcosx/tanx.
The function can be rewritten as g(x) = sec²xcosx/tanx = secxcotx. This can be differentiated using the Product rule. Let u = secx and v = cotx. Then, du/dx = secxtanx and dv/dx = ācosec²x. So, by the Product rule, g'(x) = āsecxcosec²x + cotxsecxtanx = āsecxcosec²x + secx.
3
Prove that the derivative of cotx is ācosec²x.
Let y = cotx = 1/tanx = (tanx)ā»Ā¹. So, using the Power rule, dy/dx = ā(tanx)ā»Ā²sec²x = āsec²x/tan²x = ācos²x/cos²xsin²x = ā1/sin²x = ācosec²x.
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) = ācosec²x + secxtanx.
5
Find the derivative of y = 8tan³x.
This can be done by using the Chain rule twice. Let y = 8u³ and u = tanx. Then, dy/du = 24u² and du/dx = sec²x. So, dy/dx = 24u²sec²x = 24tan²xsec²x.
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