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:(adsbygoogle = window.adsbygoogle || []).push({});

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.

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Derivatives of Trigonometric Equations

**Physics Forums | Science Articles, Homework Help, Discussion**