Determine the equation of the tangent line to the function given

[ttpp1124]

Homework Statement

The answer is supposed to be in exact value, and I'm not sure if my form is correct. Can my answer be further left in exact value, or shall I leave it as it is?

Relevant Equations

n/a

 

[Mark44]

I get a different value for the slope of f at $x = \frac{3\pi} 4$.

 

[benorin]

@Mark44 I checked his work, I also get 2:

$$\left.\tfrac{dy}{dx}\right|_{x=\tfrac{3\pi}{4}}=\left.\tfrac{d}{dx}\tan \left( 2x-\tfrac{\pi}{2}\right)\right|_{x=\tfrac{3\pi}{4}}=\left. 2\sec ^2 \left( 2x-\tfrac{\pi}{2}\right)\right|_{x=\tfrac{3\pi}{4}}=\tfrac{2}{\cos ^2 \left( \pi\right)}=2$$

 

[ttpp1124]

I get a different value for the slope of f at $x = \frac{3\pi} 4$.

sorry for the late follow-up, I checked my work again, and my slope is 2. My b value, upon calculation, is $-3\pi$. Would my final equation be $y=2x-3\pi$?

 

[Mark44]

@Mark44 I checked his work, I also get 2:

$$\left.\tfrac{dy}{dx}\right|_{x=\tfrac{3\pi}{4}}=\left.\tfrac{d}{dx}\tan \left( 2x-\tfrac{\pi}{2}\right)\right|_{x=\tfrac{3\pi}{4}}=\left. 2\sec ^2 \left( 2x-\tfrac{\pi}{2}\right)\right|_{x=\tfrac{3\pi}{4}}=\tfrac{2}{\cos ^2 \left( \pi\right)}=2$$

Right. There's a place in the image in post #1 where it says "slope = 0 + x = 3π/4", so I thought the OP was getting 0 for the slope.

Would my final equation be y=2x−3π?

Yes.

 

[ttpp1124]

Right. There's a place in the image in post #1 where it says "slope = 0 + x = 3π/4", so I thought the OP was getting 0 for the slope.
Yes.

sorry, I realized I made an error in my math upon determining the equation. It would be $y=2x-\frac{3\pi }{2}$.

 

[Mark44]

sorry, I realized I made an error in my math upon determining the equation. It would be $y=2x-\frac{3\pi }{2}$.

Yes, that's right.

Sorry for my previous incorrect answer...