Find an equation for the tangent to the curve y=1+ (sqrt2)(csc(X)) + cot(X).

I just learned, though my teacher wasn't super good with explaining, derivatives for trig functions but they still aren't making too much sense. Could you please give me a step by step solution for this example please?

Thank you so much!

so you need the derivative of this thing.

y = 1 + (cscx + cotx)^1/2

dy/dx = (1/2)(cscx + cotx)^(-1/2) ( -cscx cotx - csc^2 x)
I was going to simplify this, but then realized that you were missing some information.
e.g. what point does the tangent pass through ?

I am also puzzled by the "sqrt2" , I read it as the square root

Yes it is square root. I think I figured out the question now. Thanks!!

Sure, I'd be happy to guide you through the process of finding the equation for the tangent to the given curve.

Step 1: Start by finding the derivative of the function y with respect to x. In this case, we have y = 1 + √2 csc(x) + cot(x).

To compute the derivative, let's first differentiate each term separately using the basic rules of differentiation:

- The derivative of a constant (1 in this case) is zero.
- The derivative of √2 csc(x) can be found by using the chain rule, which states that if y = f(g(x)), then d/dx[y] = f'(g(x)) * g'(x). In this case, f(u) = √2/u and g(x) = csc(x), so the derivative of √2 csc(x) is √2 [d/dx(csc(x))] = √2 [-csc(x) cot(x)] = -√2 csc(x) cot(x).
- The derivative of cot(x) can be found using the derivative of the reciprocal function rule, which states that if y = 1/f(x), then d/dx[y] = -f'(x)/[f(x)]^2. In this case, f(x) = sin(x), so the derivative of cot(x) is -cos(x)/[sin(x)]^2 = -1/[sin(x) cos(x)].

Putting it all together, the derivative of y = 1 + √2 csc(x) + cot(x) is:
dy/dx = 0 + (-√2 csc(x) cot(x)) + (-1/[sin(x) cos(x)])

Step 2: Simplify the expression for dy/dx by combining like terms and using trigonometric identities.

dy/dx = -√2 csc(x) cot(x) - 1/[sin(x) cos(x)]

Since csc(x) = 1/sin(x), we can rewrite csc(x) as 1/[sin(x)].

dy/dx = -√2 (1/[sin(x)]) cot(x) - 1/[sin(x) cos(x)]

Step 3: Now, the equation of the tangent line can be expressed in point-slope form, y - y1 = m(x - x1), where (x1, y1) is a point on the curve and m is the slope. Since we want to find the equation of the tangent, we need to find the slope, which is equal to dy/dx evaluated at a specific point.

Let's assume we want to find the equation of the tangent at a point (a, y(a)). We can substitute x = a and solve for dy/dx:

dy/dx = -√2 (1/[sin(a)]) cot(a) - 1/[sin(a) cos(a)]

Step 4: Now, we have the slope, dy/dx, evaluated at the point (a, y(a)). We can use this point and the slope in the point-slope form to find the equation of the tangent line:

y - y(a) = dy/dx * (x - a)

Substituting the slope, we have:

y - y(a) = [-√2 (1/[sin(a)]) cot(a) - 1/[sin(a) cos(a)]] * (x - a)

This is the equation for the tangent to the curve y = 1 + √2 csc(x) + cot(x) at the point (a, y(a)).

Remember to substitute the values of a and y(a) with the specific values you want to find the equation of the tangent at.

I hope this step-by-step solution helps you understand the process of finding the equation for the tangent to a curve. Let me know if you have any further questions!