Web Page Written by Ben Mueller (Trigonometric differentiation)

It is important to remember the derivatives of the six trigonometric functions as these functions often come up in Physics and other applications of science.

Let’s review the rules for differentiating trigonometric functions.

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The derivatives of the six trigonometric functions are as follows (Y’ denotes the derivative):

If Y = Sin x, then Y’ = Cos x

If Y = Cos x, then Y’ = -Sin x

If Y = Tan x, then Y’ = (Sec x)^2

If Y = Cot x, then Y’ = -(Csc x)^2

If Y = Sec x, then Y’ = Tan x * Sec x

If Y = Csc x, then Y’ = -(Cot x * Csc x)

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Problem:

Find the derivative of the function Y = Sin (Cos x)

Solution:

By the chain rule: Y’ = Cos (Cos x) * (-Sin x)

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Problem:

Find the derivative of the function Y = Ln (Sec x + Tan x)

Solution:

By logarithmic differentiation and the chain rule: Y’ = 1/(Sec x + Tan x) * (Sec x*Tan x + (Sec x)^2). Then by factoring out the common Sec x from the top equation: Y’ = Sec x(Tan x + Sec x)/(Sec x + Tan x). We then notice that Sec x + Tan x exists on both the top and bottom half of the equation and canceling out this term yields: Y’ = Sec x. Thus the derivative Ln (Sec x + Tan x) is Sec x.

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Consider the derivatives of the following functions (Y):

If Y = Cos x * x^2, then by the product rule, Y’ = -(Sin x)*(x^2) + 2x*(Cos x) or Y’ = x(-Sin x * x + 2Cos x)

If Y = (Sin x)^3, then by the chain rule, Y’ = 3(Sin x)^2*(Cos x)

If Y = Ln (Cos x), then by logarithmic differentiation and the chain rule, Y’ = 1/(Cos x) * (-Sin x) or Y’ = -Tan x

If Y = Ln (2x) * Sec x, then by the product rule and the chain rule, Y’ = 1/2x*2*(Sec x) + (Sec x)(Tan x)(Ln 2x) or

Y’ = (Sec x)/x + (Sec x)(Tan x)(Ln 2x)

If Y = 2^(Tan x), then by the chain rule, Y’ = 2^(Tan x) * Ln 2 * (Sec x)^2.

More Calculus and Analytic Geometry Links:

• Basic Differentiation Rules (Power rule, product rule, quotient rule)
• Basic Differentiation Drill Problems
• Finding the Derivatives of Trigonometric Functions
• Differentiating Logarithmic Functions

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