Finding the equation of a tangent line at a given point (tan and e^x questions)

[czzzar]

View attachment 1213
I don't need an answer, I'm just stumped as to how to properly turn these into point slope form to find the tangent, can anyone guide me through this? Help would be greatly appreciated.

I believe I understand the formulas that are used to solve problems such as these.
It starts by finding the derivative of the function.
After finding the derivative, I can plug in the given points and slope from the derivative into the point slope formula... but when i did this, my answer was wrong. I feel like I'm missing a key idea (or more) that is stopping me from finding this out

 

[Chris L T521]

View attachment 1213

I don't need an answer, I'm just stumped as to how to properly turn these into point slope form to find the tangent, can anyone guide me through this? Help would be greatly appreciated.

I believe I understand the formulas that are used to solve problems such as these.
It starts by finding the derivative of the function.
After finding the derivative, I can plug in the given points and slope from the derivative into the point slope formula... but when i did this, my answer was wrong. I feel like I'm missing a key idea (or more) that is stopping me from finding this out

It seems to me that you have the right formula. The tangent line to $f(x)$ at $x=a$ is given by the equation $y=f(a)+f^{\prime}(a)(x-a)$. In the case of the first problem, $f(x)=\tan x$, and you're given the point $(a,f(a)) = \left(\frac{\pi}{4},1\right)$. With that, the equation of the tangent line is of the form
\[y=f\left(\frac{\pi}{4}\right) +f^{\prime}\left(\frac{\pi}{4}\right) \left(x-\frac{\pi}{4}\right) = 1 + f^{\prime}\left(\frac{\pi}{4}\right) \left(x-\frac{\pi}{4}\right).\] I'm sure you can find $f^{\prime}\left(\frac{\pi}{4}\right)$ on your own, right?

The same goes for the other question: for $f(x)=(x-1)e^x$ at the point $(1,0)$, the equation of the tangent is of the form
\[y=f(1)+f^{\prime}(1)(x-1)= f^{\prime}(1)(x-1).\]
I would assume here that you can find $f^{\prime}(1)$ as well.

I hope this clarifies things!

 

[czzzar]

It seems to me that you have the right formula. The tangent line to $f(x)$ at $x=a$ is given by the equation $y=f(a)+f^{\prime}(a)(x-a)$. In the case of the first problem, $f(x)=\tan x$, and you're given the point $(a,f(a)) = \left(\frac{\pi}{4},1\right)$. With that, the equation of the tangent line is of the form
\[y=f\left(\frac{\pi}{4}\right) +f^{\prime}\left(\frac{\pi}{4}\right) \left(x-\frac{\pi}{4}\right) = 1 + f^{\prime}\left(\frac{\pi}{4}\right) \left(x-\frac{\pi}{4}\right).\] I'm sure you can find $f^{\prime}\left(\frac{\pi}{4}\right)$ on your own, right?

The same goes for the other question: for $f(x)=(x-1)e^x$ at the point $(1,0)$, the equation of the tangent is of the form
\[y=f(1)+f^{\prime}(1)(x-1)= f^{\prime}(1)(x-1).\]
I would assume here that you can find $f^{\prime}(1)$ as well.

I hope this clarifies things!

$f^{\prime}\left(\frac{\pi}{4}\right)$ i believe is 2 because the derivative of $f(x)=\tan x$ is $f(x)=\sec^2x$...
\[y=1 + 2x - 2\frac{\pi}{4}.\] should pi/4 right be able to be multiplied? the answer to this question is 4x - 2y - pi +2 = 0... It seems that I'm deviating away from the answer even more now...

 

[Chris L T521]

$f^{\prime}\left(\frac{\pi}{4}\right)$ i believe is 2 because the derivative of $f(x)=\tan x$ is $f(x)=\sec^2x$...
\[y=1 + 2x - 2\frac{\pi}{4}.\] should pi/4 right be able to be multiplied? the answer to this question is 4x - 2y - pi +2 = 0... It seems that I'm deviating away from the answer even more now...

Yea, you should multiply them in order to simplify the equation. With that, you should have the tangent equation
\[y=2x+1-\frac{\pi}{2}.\]

However, your book goes the extra mile to get it in standard form (I don't know why they want you do get it into standard form, but what I have above is usually good enough). To get the book's answer, multiply both sides by 2 to get $2y=4x+2-\pi$ and then subtract $2y$ from both sides to get $4x-2y-\pi+2=0$.

I hope this makes sense!