Find an equation of the tangent line to the curve at the given point.

Question

y = sec x, at (3, 2)

 

I know I need to find the derivative obviously but I'm struggling with it

Answer 1

Ok you want to find the derivative of this but don't know how....

 

you know sec x =1/cos x

 

 

So lets derive 1/cos x instead of sec x

 

(Lo D hi - hi D lo)/(lo)²

 

(cos x (0)-1*(-sinx))/(cos x)²

 

So we know the derivative of sec x = sinx/(cos x)².  We can break this apart to (sinx/cosx)*1/cos x or tanx sec x. 

 

Now we can plug in 3 to find the slope.  Remember the first derivative is the slope at a point

 

you will get tan(3)sec(3).  This is the slope

 

Now use the equation of the line to find the rest

 

y-2=tan(3)sec(3)(x-3)

 

There's your answer.  But are you sure the x point is 3?????