Derivative of Trig Function | Find y' for (sec (2x))/(1 + tan (2x))

Mathematics/Calculus/CalculusThis website is a great resource for learning math and science concepts, including calculus. It provides step-by-step explanations and examples for a variety of topics, including derivatives. In summary, for the given problem, we are asked to find the derivative of y with respect to x, which is denoted as y'. After using the identity sec2x=1+tan2x, we can simplify the expression to y' = [(2 sec (2x))(tan (2x) - 1)]/[(1 + tan (2x))^2]. The website provided can be a useful tool for understanding and practicing derivative problems.
  • #1
Jumpy Lee
20
0

Homework Statement



Find y' of the following

Homework Equations



y = (sec (2x))/(1 + tan (2x))

The Attempt at a Solution



y' = [(1 + tan (2x))(2 sec (2x) tan (2x)) - (sec (2x))(2 (sec (2x))^2)]/[(1 + tan (2x))^2]


i can not quite figure out how it reduces to:

y' = [(2 sec (2x))(tan (2x) - 1)]/[(1 + tan (2x))^2] :grumpy:
 
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  • #2
Try using the identity sec2x=1+tan2x on the last term in the numerator. Then it should simplify to the result you state.
 
Last edited:
  • #3
You had

[tex]y^\prime =\frac{(1 + \tan 2x)(2\sec 2x \tan 2x) - (\sec 2x)(2 \sec^{2} 2x)}{(1 + \tan 2x)^2}=\frac{2\sec 2x \left[(1 + \tan 2x)\tan 2x - \sec^{2} 2x\right]}{(1 + \tan 2x)^2}[/tex]
[tex]=\frac{2\sec 2x \left[\tan 2x + \tan^{2} 2x - (1 + \tan^{2} 2x)\right]}{(1 + \tan 2x)^2}=\frac{2\sec 2x (\tan 2x - 1)}{(1 + \tan 2x)^2}[/tex]
 
  • #4
thank you cristo and benorin
 

What is the derivative of a trigonometric function?

The derivative of a trigonometric function is the rate of change of the function at a particular point. It represents the slope of the tangent line to the function at that point.

How do you find the derivative of a trigonometric function?

The derivative of a trigonometric function can be found using the standard rules of differentiation, such as the power rule, product rule, and chain rule. The specific rules used will depend on the form of the trigonometric function.

What are the common trigonometric functions?

The most commonly used trigonometric functions are sine, cosine, and tangent. Other common trigonometric functions include secant, cosecant, and cotangent. All of these functions are related to the sides and angles of a right triangle.

Why are derivatives of trigonometric functions important?

Derivatives of trigonometric functions are important in many branches of science and engineering, such as physics, astronomy, and calculus. They allow us to model and analyze real-world phenomena involving periodic motion and oscillations.

How can derivatives of trigonometric functions be applied in real life?

Derivatives of trigonometric functions have numerous applications in real life, including in fields such as engineering, physics, and economics. They are used to optimize processes, analyze data, and solve problems in a variety of real-world situations.

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