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### Course: Calculus, all content (2017 edition) > Unit 2

Lesson 26: Trigonometric functions differentiation- Derivatives of tan(x) and cot(x)
- Derivatives of sec(x) and csc(x)
- Derivatives of tan(x), cot(x), sec(x), and csc(x)
- Worked example: Derivative of sec(3π/2-x) using the chain rule
- Differentiate trigonometric functions
- Derivative of tan(x) (old)
- Differentiating trigonometric functions review

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# Differentiating trigonometric functions review

Review your trigonometric function differentiation skills and use them to solve problems.

## How do I differentiate trigonometric functions?

First, you should know the derivatives for the basic trigonometric functions:

You can actually use the derivatives of sine and cosine (along with the quotient rule) to obtain the derivatives of all the other functions.

*Want to learn more about differentiating trigonometric functions? Check out this video about sine and cosine, this video about tangent and cotangent, and this video about secant and cosecant.*

## Practice set 2: tangent, cotangent, secant, and cosecant

*Want to try more problems like this? Check out this exercise.*

After you've mastered the derivatives of the basic trigonometric functions, you can differentiate trigonometric functions whose arguments are polynomials, like $\mathrm{sec}({\displaystyle \frac{3\pi}{2}}-x)$ .

## Practice set 3: general trigonometric functions

*Want to try more problems like this? Check out this exercise.*

## Want to join the conversation?

- In the video "Derivatives of sec(x) and csc(x)", you defined the derivative of csc(x) as -cot(x)*csc(x); here it's listed as -csc(x)*cot(x). Does it matter?(3 votes)
- No. Multiplication is commutative. -csc(x)*cot(x) = -cot(x)*csc(x)(5 votes)

- For dr/dx tan (x), I'm struggling with the quotient rule. Why are we not putting sin^2 (x ) in the denominator? Sinx/cosx -> (cosx/cosx) + (sinx/sin^x) then combine. Im trying to work through the quotient rule rather than jump to the (cos^2 + sin^2)/cos^2. Thank you so much(1 vote)
- tan(x) = sin(x)/cos(x) as you noted.

Let f(x) = sin(x) and g(x) = cos(x).

This means f'(x) = cos(x) and g'(x) = -sin(x).

The the quotient rule is structured as [f'(x)*g(x) - f(x)*g'(x)] / g(x)^2.

In your question above you noted that the terms should be divided and that is not the case as they should be multiplied together.

If we sub in terms to the quotient rule (being careful to keep track of signs) we get the following:

[cos(x)*cos(x) - (sin(x)*-sin(x))]/cos^2(x)

[cos^2(x) + sin^2(x)]/cos^2(x)

With the trig identity we know cos^2(x) + sin^2(x) = 1 therefore with our final substitution we get...

1/cos^2(x) aka sec^2(x)

Hope this helps!(4 votes)

- Differential of 5 cos x-2(0 votes)
- What is the derivative of sin2x^3(1 vote)
- How are we going to differentiate sec^3(x/2)(0 votes)
- Here we have to apply the product rule twice:

d/dx(sec^3(x/2))

=d/dx(sec(x/2)*sec^2(x/2)

=[sec^2(x/2)*d/dx{sec(x/2) } ] + [sec(x/2)*d/dx{sec^2(x/2) } ]

=[sec^2(x/2)*sec(x/2)*tan(x/2)*d/dx(x/2)] +

[sec(x/2)*{ (sec(x/2)*d/dx(sec(x/2)))+(sec(x/2)*d/dx(sec(x/2)))}]

=[1/2(sec^3(x/2)*tan(x/2))] + [2*sec^2(x/2)*d/dx(sec(x/2))]

=[1/2(sec^3(x/2)*tan(x/2))] + [2*sec^2(x/2)*sec(x/2)*tan(x/2)*d/dx(x/2)]

=[1/2(sec^3(x/2)*tan(x/2))] + [(2/2)*(sec^3(x/2)tan(x/2))]

=3/2[sec^3(x/2)*tan(x/2)](1 vote)