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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:
start fraction, d, divided by, d, x, end fraction, sine, left parenthesis, x, right parenthesis, equals, cosine, left parenthesis, x, right parenthesis
start fraction, d, divided by, d, x, end fraction, cosine, left parenthesis, x, right parenthesis, equals, minus, sine, left parenthesis, x, right parenthesis
start fraction, d, divided by, d, x, end fraction, tangent, left parenthesis, x, right parenthesis, equals, \sec, squared, left parenthesis, x, right parenthesis, equals, start fraction, 1, divided by, cosine, squared, left parenthesis, x, right parenthesis, end fraction
start fraction, d, divided by, d, x, end fraction, cotangent, left parenthesis, x, right parenthesis, equals, minus, \csc, squared, left parenthesis, x, right parenthesis, equals, minus, start fraction, 1, divided by, sine, squared, left parenthesis, x, right parenthesis, end fraction
start fraction, d, divided by, d, x, end fraction, \sec, left parenthesis, x, right parenthesis, equals, \sec, left parenthesis, x, right parenthesis, tangent, left parenthesis, x, right parenthesis, equals, start fraction, sine, left parenthesis, x, right parenthesis, divided by, cosine, squared, left parenthesis, x, right parenthesis, end fraction
start fraction, d, divided by, d, x, end fraction, \csc, left parenthesis, x, right parenthesis, equals, minus, \csc, left parenthesis, x, right parenthesis, cotangent, left parenthesis, x, right parenthesis, equals, minus, start fraction, cosine, left parenthesis, x, right parenthesis, divided by, sine, squared, left parenthesis, x, right parenthesis, end fraction
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 1: sine and cosine

Problem 1.1
f, left parenthesis, x, right parenthesis, equals, 2, x, minus, 3, sine, left parenthesis, x, right parenthesis
f, prime, left parenthesis, x, right parenthesis, equals

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

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

Problem 2.1
Let f, left parenthesis, x, right parenthesis, equals, tangent, left parenthesis, x, right parenthesis.
Find f, prime, left parenthesis, start fraction, pi, divided by, 6, end fraction, right parenthesis.
Choose 1 answer:
Choose 1 answer:

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 \sec, left parenthesis, start fraction, 3, pi, divided by, 2, end fraction, minus, x, right parenthesis.

Practice set 3: general trigonometric functions

Problem 3.1
g, left parenthesis, x, right parenthesis, equals, sine, left parenthesis, 4, x, squared, plus, 3, x, right parenthesis
g, prime, left parenthesis, x, right parenthesis, equals, question mark
Choose 1 answer:
Choose 1 answer:

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

Want to join the conversation?

  • starky ultimate style avatar for user Jesse
    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?
    (2 votes)
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  • blobby green style avatar for user scott albert
    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)
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    • old spice man green style avatar for user Eastman Landry
      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)
  • blobby green style avatar for user kawalker310
    Differential of 5 cos x-2
    (0 votes)
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  • blobby green style avatar for user maria.villafuerte
    What is the derivative of sin2x^3
    (1 vote)
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  • mr pink red style avatar for user Aashutosh Dubey
    How are we going to differentiate sec^3(x/2)
    (0 votes)
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    • duskpin ultimate style avatar for user shrivatsa mishra
      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)