Differentiation of trigonometric functions for OCR A-level Maths

Differentiation of trigonometric functions

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 = −sinx and d(tanx)/dx = sec²x.

Differentiating trigonometric functions

Equations involving trigonometric reciprocals can be differentiated using the basic rules of differentiation by using the following results: d(cotx)/dx = −cosec²x, d(secx)/dx = secxtanx and d(cosecx)/dx = −cosecxcotx.

Differentiating trigonometric reciprocals

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

Differentiating trigonometric functions with the chain rule

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

Differentiating trigonometric functions without the chain rule

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 = x⁻¹ and v = sinx. Then, du/dx = −1/x² and dv/dx = cosx. So, by the Product rule, dy/dx = −sinx/x² + cosx/x.

Find a function for the slope for the given graph.

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 = −cosec²x. So, by the Product rule, g'(x) = −secxcosec²x + cotxsecxtanx = −secxcosec²x + secx.

Differentiate the following function: g(x) = sec²xcosx/tanx.

3

Prove that the derivative of cotx is −cosec²x.

Let y = cotx = 1/tanx = (tanx)⁻¹. So, using the Power rule, dy/dx = −(tanx)⁻²sec²x = −sec²x/tan²x = −cos²x/cos²xsin²x = −1/sin²x = −cosec²x.

Prove that the derivative of cotx is −cosec²x.

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) = −cosec²x + secxtanx.

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

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.

Find the derivative of y = 8tan³x.

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