Equation of the tangent line to y = tan(x)

[SCGirl]

Hi, this problem seemed easy, but I just can't get the right answer. Here it is:

Find the equation of the tangent line to the curve at the given point. The curve is y = tanx, and the point is (pi/4, 1).

I took the derivative and got (secx)^2. Plugging in the point, I got 1 as the slope and my equation is this: y = 1+ (x-(pi/4)). The answer should be: y = 2x + 1 - (pi/2). Please help.

 

[galactus]

You just have some minor mistakes.



Your derivative is correct: \(\displaystyle sec^{2}(x)\)

Using \(\displaystyle \L\\y=mx+b\):

\(\displaystyle \L\\\overbrace{1}^{\text{y}}=\overbrace{sec^{2}(\frac{\pi}{4})}^{\text{m}}\overbrace{(\frac{\pi}{4})}^{\text{x}}+b\)

Solving for b gives \(\displaystyle \L\\b=1-\frac{\pi}{2}\)

Therefore, since \(\displaystyle \L\\sec^{2}(\frac{\pi}{4})=2\) your equation is:

\(\displaystyle \L\\y=2x+(1-\frac{\pi}{2})\)

 

[SCGirl]

Thank you!