Derivatives of Trigonometric Equations

In summary, the conversation discussed finding the equation of the tangent line to a curve at a given point using the derivative. The first problem involved the function y = 1/(sinx+cosx) and after solving for the derivative, the equation of the tangent line was found to be y = -x + 1. In the second problem, the function was y = sqrt(x^2+5) and the derivative was correctly calculated as x / sqrt(x^2+5), resulting in the equation of the tangent line y = 2/3x + 5/3. The conversation also addressed a mistake in the calculation of the derivative for the first problem and clarified the steps for finding the derivative.
  • #1
loadsy
57
0
Alright, moving on to the topic of taking the derivative of a trigonometric equations, I have been given the problem to find the equation of the tangent line to the curve at the given point for:

y = 1/sinx+cosx at (0,1)

Now we know that the equation is y-1=m(x-0), so I've tried solving for the derivative and got as far as this:
y' = (sinx +cosx)d/dx(1) - 1 d/dx (sinx+cosx)
-------------------------------------
(sinx+cosx)^2

= (sinx + cosx)X1 - (1 X (cosx + -sinx) )
-------------------------------
(sinx + cosx)^2


= 2sinx / (sinx + cosx)^2 = 2 cosx / (sinx+cosx)^2

And then subbing in x=0 and getting 1/1 = 1 and then the equation would be y=x+1, however, I'm not sure if I did this right, or if there is some other way to simplify it that I may have misinterpreted.

Also, while we are at it, I've just done another question that asks to "find the tangent line to the curve of:"

y = squareroot(X^2+5) at (2,3)

I went ahead and solved the derivative and found it to be x/(squareroot(X^2+5)) and subbing in the x=2 got 2/7 as my answer and then just put it into the equation to get y = 2/7X + 17/7, is this correct?

Thank you anyone, I apologize for asking two questions in one topic haha.
 
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  • #2
[tex] y = \frac{1}{\sin x + \cos x} [/tex].

[tex] y' = -\frac{1(\cos x - \sin x)}{(\sin x + \cos x)^{2}} [/tex]. It should be [tex] y = -x +1 [/tex][tex] y' = \frac{x}{\sqrt{x^{2}+5}} [/tex]. You should get [tex] m = \frac{2}{3} [/tex] not [tex] \frac{2}{7} [/tex].
 
Last edited:
  • #3
loadsy said:
Alright, moving on to the topic of taking the derivative of a trigonometric equations, I have been given the problem to find the equation of the tangent line to the curve at the given point for:

y = 1/sinx+cosx at (0,1)

Now we know that the equation is y-1=m(x-0), so I've tried solving for the derivative and got as far as this:
y' = (sinx +cosx)d/dx(1) - 1 d/dx (sinx+cosx)
-------------------------------------
(sinx+cosx)^2

= (sinx + cosx)X1 - (1 X (cosx + -sinx) )
-------------------------------
(sinx + cosx)^2


= 2sinx / (sinx + cosx)^2 = 2 cosx / (sinx+cosx)^2

And then subbing in x=0 and getting 1/1 = 1 and then the equation would be y=x+1, however, I'm not sure if I did this right, or if there is some other way to simplify it that I may have misinterpreted.

Also, while we are at it, I've just done another question that asks to "find the tangent line to the curve of:"

y = squareroot(X^2+5) at (2,3)

I went ahead and solved the derivative and found it to be x/(squareroot(X^2+5)) and subbing in the x=2 got 2/7 as my answer and then just put it into the equation to get y = 2/7X + 17/7, is this correct?

Thank you anyone, I apologize for asking two questions in one topic haha.

First, yout function was probably y(x) = 1 / (sinx + cosx). You should watch out for the brackets. Second, what is the derivatice of a constant (in your case 1)?

Regarding the second function, your answer seems to be ok.
 
  • #4
For the second function, the square root of 9 is 3, not 7
 
  • #5
Right, the equation of the tangent line should be y = 2/3 x + 5/3.
 
  • #6
Yeah I realized that mistake after courtrigrad posted it. I ended up not squarerooting the 5 in X^2+5. So the answer would be 2/3 and it ends up being y= 2/3X + 5/3. I get that question, however, can you explain to me in the 1st problem, why it is -1(cosx-sinx)/(sinx+cosx)^2 because I'm currently going back and reworking that problem out and I'm not sure how you get the -1 part of it, unless I did a step in there that isn't jiving with everything else.
 
  • #7
loadsy said:
Yeah I realized that mistake after courtrigrad posted it. I ended up not squarerooting the 5 in X^2+5. So the answer would be 2/3 and it ends up being y= 2/3X + 5/3. I get that question, however, can you explain to me in the 1st problem, why it is -1(cosx-sinx)/(sinx+cosx)^2 because I'm currently going back and reworking that problem out and I'm not sure how you get the -1 part of it, unless I did a step in there that isn't jiving with everything else.

Your function is [tex]f(x) = \frac{1}{\sin x + \cos x} = \frac{g(x)}{h(x)}[/tex]. So, [tex]f'(x) = \frac{g'(x) h(x) - g(x)h'(x)}{h(x)^2}[/tex]. Plug in and solve.
 
  • #8
[tex] y = \frac{1}{\sin x + \cos x} [/tex]

[tex] f(x) = 1 [/tex]
[tex] g(x) = \sin x + \cos x [/tex].

[tex] f'(x) = 0 [/tex]
[tex] g'(x) = \cos x - \sin x [/tex].

[tex] \frac{dy}{dx} = \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^{2}} [/tex]
[tex] \frac{dy}{dx} = \frac{(\sin x + \cos x)(0)-1(\cos x-\sin x)}{(\sin x + \cos x)^{2}} [/tex]
 
  • #9
Oh my goodness, alright I'm following you now, the problem lied within the fact that f'(x) was equal to 0, so I still had the (sinx + cosx) still in the equation. Alright, thanks for your time all of you. I think I'm ready to go back and do some more practice problems to help me grasp this concept fully. :D
 

FAQ: Derivatives of Trigonometric Equations

What are the basic trigonometric derivatives?

The basic trigonometric derivatives are:

  • sin(x) = cos(x)
  • cos(x) = -sin(x)
  • tan(x) = sec^2(x)
  • cot(x) = -csc^2(x)
  • sec(x) = sec(x)tan(x)
  • csc(x) = -csc(x)cot(x)

Can the chain rule be applied to trigonometric derivatives?

Yes, the chain rule can be applied to trigonometric derivatives. For example, if the function is sin(2x), then the derivative would be cos(2x) * 2, since the derivative of 2x is 2 and the derivative of sin(x) is cos(x).

How do you find the derivative of inverse trigonometric functions?

To find the derivative of inverse trigonometric functions, you can use the following formulas:

  • d/dx(arcsin(x)) = 1 / sqrt(1 - x^2)
  • d/dx(arccos(x)) = -1 / sqrt(1 - x^2)
  • d/dx(arctan(x)) = 1 / (1 + x^2)
  • d/dx(arccot(x)) = -1 / (1 + x^2)
  • d/dx(arcsec(x)) = 1 / (x * sqrt(x^2 - 1))
  • d/dx(arccsc(x)) = -1 / (x * sqrt(x^2 - 1))

Can trigonometric derivatives be used to solve real-world problems?

Yes, trigonometric derivatives can be used to solve real-world problems, particularly in physics and engineering. For example, they can be used to calculate the velocity or acceleration of an object in motion, or to determine the slope of a ramp or hill.

Are there any common mistakes to avoid when finding derivatives of trigonometric equations?

One common mistake to avoid is forgetting to use the chain rule when the variable inside the trigonometric function is not just x. Another mistake is not simplifying the derivative by using trigonometric identities. It is also important to remember that the derivative of a trigonometric function is not always just the opposite trigonometric function, as shown in the basic trigonometric derivatives. Finally, always check your answers by plugging them back into the original equation to ensure they are correct.

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