# Derivatives of Trigonometric Equations

1. Oct 15, 2006

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.

2. Oct 15, 2006

$$y = \frac{1}{\sin x + \cos x}$$.

$$y' = -\frac{1(\cos x - \sin x)}{(\sin x + \cos x)^{2}}$$. It should be $$y = -x +1$$

$$y' = \frac{x}{\sqrt{x^{2}+5}}$$. You should get $$m = \frac{2}{3}$$ not $$\frac{2}{7}$$.

Last edited: Oct 15, 2006
3. Oct 15, 2006

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)?

4. Oct 15, 2006

### Office_Shredder

Staff Emeritus
For the second function, the square root of 9 is 3, not 7

5. Oct 15, 2006

Right, the equation of the tangent line should be y = 2/3 x + 5/3.

6. Oct 15, 2006

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. Oct 15, 2006

Your function is $$f(x) = \frac{1}{\sin x + \cos x} = \frac{g(x)}{h(x)}$$. So, $$f'(x) = \frac{g'(x) h(x) - g(x)h'(x)}{h(x)^2}$$. Plug in and solve.

8. Oct 15, 2006

$$y = \frac{1}{\sin x + \cos x}$$

$$f(x) = 1$$
$$g(x) = \sin x + \cos x$$.

$$f'(x) = 0$$
$$g'(x) = \cos x - \sin x$$.

$$\frac{dy}{dx} = \frac{g(x)f'(x)-f(x)g'(x)}{g(x)^{2}}$$
$$\frac{dy}{dx} = \frac{(\sin x + \cos x)(0)-1(\cos x-\sin x)}{(\sin x + \cos x)^{2}}$$

9. Oct 15, 2006