Derivative of Trigonometric Functions

In summary: Can you please clarify what happened there?I think I made a mistake and it should be 2/(2√3+1).Please enter a valid response.In summary, the student attempted to solve a homework equation using the quotient rule and found that the derivatives were incorrect. He then attempted to simplify the equation by changing the terms in the denominator. After making the change, he was not able to solve the equation anymore.
  • #1
thatguythere
91
0

Homework Statement


d/dx(sec(x)/1+tan(x)
Evaluate at x=∏/6

Homework Equations





The Attempt at a Solution


((1/cos(x))(1+tan(x))-(sec(x))(sin(x)/cos(x)))/(1+tan(x))^2

((1/cos(x))-(sec(x))(sin(x)/cos(x)))/(1+tan(x))

I used the quotient rule and reduced what I could. Have I done this right and can I go any further before evaluating? Thank you.
 
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  • #2
No, that's not correct. Applying the quotient rule involves taking some derivatives. It doesn't appear you've done that.
 
  • #3
Of course I have. 1/cos(x) is the derivative of sec(x) and (sin(x)/cos(x)) is the derivative of (1+tan(x)) are they not?
 
  • #4
No, they're not.
 
  • #5
Right, they are not. Hahaha. Don't mind my red cheeks.
 
  • #6
=((1+tanx)(secxtanx)-(secx)(sec^2x))/(1+tanx)2
=(secxtanx+secxtan^2x-sec^3x)/(1+tanx)2

Is this looking any closer to a correct answer?
 
  • #7
Yup!
 
  • #8
Is this as far as it goes? Can it be simplified any more?
 
  • #9
thatguythere said:
Of course I have. 1/cos(x) is the derivative of sec(x) and (sin(x)/cos(x)) is the derivative of (1+tan(x)) are they not?

I realize that you already recognized that these are wrong, but I thought it worth mentioning why they were wrong.

sec(x) = 1/cos(x) - This is an identity and is how the secant function is defined. It has nothing to do with derivatives.

tan(x) = sin(x)/cos(x) - This is also an identity and is how the tangent function is defined. This is unrelated to derivatives.
 
  • #10
Yes, I was simply confusing myself. As far as my current work, can it be simplified any more?
 
  • #11
thatguythere said:
Yes, I was simply confusing myself. As far as my current work, can it be simplified any more?
You might be able to, but what's the point? Just go ahead and evaluate it now at x = ##\pi/6##.
 
  • #12
((2/√3)(√3/3)+((2/√3)(1/3))-(8/3^(3/2)))/(4/3)+((2√3)/3)
=(2/9+2/(3√3)-8/(3^3/2))/((2(2√3+1))/3√3)
 
  • #13
Can anyone tell me if I am on the right track please?
 
  • #14
thatguythere said:
Can anyone tell me if I am on the right track please?

thatguythere said:
=((1+tanx)(secxtanx)-(secx)(sec^2x))/(1+tanx)2
=(secxtanx+secxtan^2x-sec^3x)/(1+tanx)2
The above is fine, but what you have below is difficult to read, especially what you have on the right side of the =. Please simplify it.


thatguythere said:
((2/√3)(√3/3)+((2/√3)(1/3))-(8/3^(3/2)))/(4/3)+((2√3)/3)
=(2/9+2/(3√3)-8/(3^3/2))/((2(2√3+1))/3√3)
 
  • #15
Let's try this
 

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  • #16
Oh I guess I could change the term in the denominator to 4/3.
 
  • #17
Bump.
 
  • #18
In the 3rd line of your attachment, the sign of the last term mysteriously changed.
 

1. What is the derivative of sine?

The derivative of sine is cosine. This means that the rate of change of the sine function at any given point is equal to the value of the cosine function at that same point.

2. How do you find the derivative of tangent?

The derivative of tangent can be found by using the quotient rule, which states that the derivative of a fraction is equal to the denominator times the derivative of the numerator minus the numerator times the derivative of the denominator, all divided by the square of the denominator.

3. Can you use the chain rule to find the derivative of a trigonometric function?

Yes, the chain rule can be used to find the derivative of a trigonometric function. The chain rule states that the derivative of a composition of functions is equal to the derivative of the outer function evaluated at the inner function, multiplied by the derivative of the inner function.

4. What is the derivative of inverse trigonometric functions?

The derivative of inverse trigonometric functions can be found by using the inverse function theorem, which states that the derivative of an inverse function is equal to the reciprocal of the derivative of the original function evaluated at the inverse function.

5. How do you find the derivative of secant and cosecant?

The derivatives of secant and cosecant can be found by using the quotient rule, similar to the derivative of tangent. However, since secant and cosecant are not fractions, the quotient rule must be modified by multiplying by the derivative of the denominator. The derivative of secant is equal to secant times tangent, and the derivative of cosecant is equal to -cosecant times cotangent.

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