Question

Help with this
Find the equation of the tangent line
y=tanx at (pi/4,1) verify graphically

Answer 1

find the 1st derivative, that gives you the equation of the slope

then substitute pi/4 to find the slope.

Now you can find the equation, you have the slope and point

hope it makes sense.

Answer 2

so the first derivative is -sinx

Answer 3

wait no. My bad i meant it's sec^2x right?

Answer 4

yes so substitute x = pi/4 to find the slope

sec = 1/cos

so the slope is

\[\sec^2(\frac{\pi}{4})=(\sec(\frac{\pi}{4}) )^2= (1/\cos(\pi/4))^2 = (\frac{\sqrt{2}}{1})^2\]

Answer 5

okay i understand everything except how u got (sqrt2/1)^2

Answer 6

ok you are working with an exact value



\[\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}\]

since

sec = 1/cos

take the reciprocal of the exact value of sec(pi/4)

\[\sec(\frac{\pi}{4})= \frac{1}{\cos(\frac{\pi}{4})} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2} \]