Derivative of a trigonometric function

In summary: So ##\ \cos(\frac{π}{10}x) = 0 \ ##. If ##\ \cos(\frac{π}{10}x) = 0 \ ##, then ##\ \frac{π}{10}x \ ## must be ##\ \frac{π}{2} \ ##, since that is the only zero of ##\ \cos(x) \ ## on the interval ##\ [0, 2π] \ ##. So ##\ \frac{π}{10}x = \frac{π}{2} \ ## and ##\ x = 5 \ ##.
  • #1
Ocata
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5

Homework Statement



[itex]\frac{d}{dx}7.5sin(\frac{pi}{10}x) [/itex]

The Attempt at a Solution



[itex]7.5(\frac{pi}{10})cos(\frac{pi}{10}x)[/itex]

Maximum: f'(x) = 0

[itex]7.5(\frac{pi}{10})cos(\frac{pi}{10}x)[/itex] = 0

[itex]7.5(\frac{pi}{10})cos^{-1}(0)= \frac{pi}{10}x[/itex]

**[itex] (\frac{pi}{10}\frac{10}{pi})7.5(90) = x[/itex]

[itex](1)(7.5)(90) = x = 675[/itex]

To me, this doesn't seem to be nearly the correct answer because it doesn't make sense given the graph of this function:

[itex]\frac{pi}{10}x= pi[/itex]

x = 10

So, the first arch is at x=0 and x = 10,

so the the maximum of the curve can not be x = 675.

What am I doing incorrectly in the derivative of the trigonometric function?

Thank you
 
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  • #2
Ocata said:

Homework Statement



[itex]\frac{d}{dx}7.5sin(\frac{pi}{10}x) [/itex]

The Attempt at a Solution



[itex]7.5(\frac{pi}{10})cos(\frac{pi}{10}x)[/itex]

Maximum: f'(x) = 0

[itex]7.5(\frac{pi}{10})cos(\frac{pi}{10}x)[/itex] = 0

[itex]7.5(\frac{pi}{10})cos^{-1}(0)= \frac{pi}{10}x[/itex]

**[itex] (\frac{pi}{10}\frac{10}{pi})7.5(90) = x[/itex]

[itex](1)(7.5)(90) = x = 675[/itex]

To me, this doesn't seem to be nearly the correct answer because it doesn't make sense given the graph of this function:

[itex]\frac{pi}{10}x= pi[/itex]

x = 10

So, the first arch is at x=0 and x = 10,

so the the maximum of the curve can not be x = 675.

What am I doing incorrectly in the derivative of the trigonometric function?

Thank you
Derivation part looks fine.
 
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  • #3
Ocata said:

Homework Statement



[itex]\frac{d}{dx}7.5sin(\frac{pi}{10}x) [/itex]
What, exactly, was the question? You only mention differentiating the function but then set that derivative to 0.

The Attempt at a Solution



[itex]7.5(\frac{pi}{10})cos(\frac{pi}{10}x)[/itex]
Yes, this is the correct derivative.

Maximum: f'(x) = 0

[itex]7.5(\frac{pi}{10})cos(\frac{pi}{10}x)[/itex] = 0
and so [itex]cos(\frac{\pi}{10}x)= 0[/itex]

[itex]7.5(\frac{pi}{10})cos^{-1}(0)= \frac{pi}{10}x[/itex]
No. To solve [itex]Af(x)= B[/itex], you take [itex]f^{-1}(B/A)[/itex], not [itex]Af^{-1}(B)[/itex].

**[itex] (\frac{pi}{10}\frac{10}{pi})7.5(90) = x[/itex]/quote]

[itex](1)(7.5)(90) = x = 675[/itex]

To me, this doesn't seem to be nearly the correct answer because it doesn't make sense given the graph of this function:

[itex]\frac{pi}{10}x= pi[/itex]

x = 10

So, the first arch is at x=0 and x = 10,

so the the maximum of the curve can not be x = 675.

What am I doing incorrectly in the derivative of the trigonometric function?

Thank you
 
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  • #4
lep11 said:
Derivation part looks fine.

Hi lep11,

Thank you. I had a feeling I was running into the issue at solving for x.
(1) [itex]7.5\frac{π}{10}cos(\frac{π}{10}x)=0[/itex]

(2) [itex]7.5\frac{π}{10}cos^{-1}(0)= \frac{π}{10}x[/itex] Here is where I was getting stuck. The first thing I noticed, was that I let cos(0) = 90 instead of π/2.

(3) [itex] 7.5\frac{π}{10}\frac{10}{π}cos(0) = x[/itex]

(4) [itex]7.5\frac{π}{2} = x = 11.78[/itex] Which is closer to within the interval of the first arch of the function, but not quite where the maximum should be.

Playing around with the algebra, I finally arrived at an answer that makes sense:

If I divide the [itex]7.5\frac{π}{10}[/itex] out of the equation first at step (1):

[itex]7.5\frac{π}{10}cos(\frac{π}{10}x)=0[/itex]

[itex]cos(\frac{π}{10}x) = 0[/itex]

[itex]cos(0) = \frac{π}{10}x[/itex]

[itex]\frac{10}{π}\frac{π}{2} = 5 [/itex] That sounds more like it.However, what I don't understand, is how come I have to divide [itex]7.5\frac{π}{10}[/itex] first? In a regular equation, it doesn't matter when you decide to divide both sides by a number, why does it matter in this situation?

Thank you.
 
Last edited:
  • #5
HallsofIvy said:
What, exactly, was the question? You only mention differentiating the function but then set that derivative to 0. No. To solve [itex]Af(x)= B[/itex], you take [itex]f^{-1}(B/A)[/itex], not [itex]Af^{-1}(B)[/itex].
Thank you HallsofIvy,

[itex]7.5\frac{π }{10}Cos(\frac{π}{10}x) = 0[/itex]

[itex]Cos(\frac{0}{(7.5\frac{π }{10})}) = \frac{π}{10}x[/itex]

[itex]Cos(0) = \frac{π}{10}x[/itex]

[itex]\frac{10}{π}Cos(0) = x[/itex]

[itex]\frac{10}{π}Cos(0) = x[/itex]

[itex]\frac{10}{π}\frac{π}{2} = x = 5[/itex]!Thank you. Now that I know how to solve this and that there is a systematic way to approach, I will be looking into why the formula provided does in fact work. Thank you for your guidance :)
 
  • #6
Ocata said:
Thank you HallsofIvy,

[itex]7.5\frac{π }{10}Cos(\frac{π}{10}x) = 0[/itex]
Thank you. Now that I know how to solve this and that there is a systematic way to approach, I will be looking into why the formula provided does in fact work. Thank you for your guidance :)
The product of ##\ 7.5\frac{π }{10} \ ## and ##\ \cos(\frac{π}{10}x) \ ## is zero.The only way for a product to be zero is for one of the factors to be zero. The only one which can be zero is ##\ \cos(\frac{π}{10}x) \ ## .
 
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What is the derivative of a trigonometric function?

The derivative of a trigonometric function is the rate of change of the function at a specific point. It measures how much the function is changing at that point.

What are the basic trigonometric functions?

The basic trigonometric functions are sine, cosine, tangent, cotangent, secant, and cosecant. These functions relate the angles and sides of a right triangle.

How do you find the derivative of a trigonometric function?

To find the derivative of a trigonometric function, you can use the chain rule or memorize the derivative formulas for each individual trigonometric function. The derivative of sine is cosine, the derivative of cosine is negative sine, and the derivative of tangent is secant squared.

Why is the derivative of a trigonometric function important?

The derivative of a trigonometric function is important because it allows us to analyze the behavior of the function, such as finding critical points, determining concavity, and finding the slope of the function at a given point. It also has many real-world applications in fields like physics, engineering, and economics.

What are some common mistakes when finding the derivative of a trigonometric function?

Some common mistakes when finding the derivative of a trigonometric function include forgetting to use the chain rule, making errors in the derivative formulas for each function, and forgetting to simplify the derivative before plugging in values. It is important to carefully follow the steps and check your work for accuracy.

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