Thread: Find the equation of the tangent line to the curve.

Hi guys,

Could someone take a look at my work.. im lost.

Problem:

Code:

Find the equation of the tangent line to the curve  
y= 4 tan x  at the point ( pi/4 , 4). 

The equation of this tangent line can be written in the form y = mx+b

my work..

Code:

using y-y1 = m(x-x1)

y - 4tan(x) = 4tan(x- (pi/4))
y - 4tan(x) = 4tan(x) - (pi tan (x))
y = 8tan(x) - pi tan(x)

m = 8tan(x)
b = pi tan (x)

Where did I go wrong? How do I get the equation from the point and the tangent line?

is it the answer?

Originally Posted by mant1s

Hi guys,

Could someone take a look at my work.. im lost.

Problem:

Code:

Find the equation of the tangent line to the curve  
y= 4 tan x  at the point ( pi/4 , 4). 

The equation of this tangent line can be written in the form y = mx+b

my work..

Code:

using y-y1 = m(x-x1)

y - 4tan(x) = 4tan(x- (pi/4))
y - 4tan(x) = 4tan(x) - (pi tan (x))
y = 8tan(x) - pi tan(x)

m = 8tan(x)
b = pi tan (x)

Where did I go wrong? How do I get the equation from the point and the tangent line?

The function is . To find the slope for any x value, differentiate it. We now see that . Since we're looking for the eqn of a tangent at , we want to find the slope of the function at . So it follows that . This is your slope in your tangent line equation.

So using the point-slope equation, we have

Does this make sense?

Rachel and Chris,

You guys rock! It makes sense to me now. I wasn't differentiating and it was screwing me up royally. Thanks for your time!

-M