Equation of tangent line to f(x) = tan x at point (pi/4, 1)
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pka
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Re: Equation of tangent line
The derivative of a function is the slope of the graph at any point, \(\displaystyle \left( {a,f(a)} \right)\).
The equation of the tangent is \(\displaystyle y = f'(a)\left( {x - a} \right) + f(a)\).
I can tell you that because you had to ask this question, you are off to a bad start in calculus. You had best really learn these basic ideas. If you don’t there is no hope. It only gets harder and more abstract.
That is the fundamental definition of derivative.yeloc said:
The derivative of a function is the slope of the graph at any point, \(\displaystyle \left( {a,f(a)} \right)\).
The equation of the tangent is \(\displaystyle y = f'(a)\left( {x - a} \right) + f(a)\).
I can tell you that because you had to ask this question, you are off to a bad start in calculus. You had best really learn these basic ideas. If you don’t there is no hope. It only gets harder and more abstract.
I am in the PROCESS of learning this material. And I am definitely not questioning your advice; I am questioning why you would belittle someone for not understanding a problem. I know that the derivative of a function is the slope of the graph at a specific point. The question I had was how to find the tangent line at a specific point.
It's not like I am asking you to work the problem out and give me the answer. I was asking for someone to guide me in the right direction of how to work the problem.
It's not like I am asking you to work the problem out and give me the answer. I was asking for someone to guide me in the right direction of how to work the problem.
pka
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You have an enormous chip on you shoulders.yeloc said:
If you took my remarks a belittlement, then you are going to have a very unpleasant life ahead of you. I am not against you; most other people are not against you. Honest advice, even if it is negative, is not meant as a put down. If you choose to attend a good college and you do not change that attitude, then you are going have a very unpleasant experience.
I have a chip on my shoulders?
I ask a question and you tell me that I am going to have a hard time in calculus? I don't need your opinions. If you have no interest in answering my question then please quit replying because you are keeping this thread off-topic. Thanks!
I ask a question and you tell me that I am going to have a hard time in calculus? I don't need your opinions. If you have no interest in answering my question then please quit replying because you are keeping this thread off-topic. Thanks!
Ok. I plugged ?/4 into the derivative of f (x)= tan x, and got 2. Then I used the slope intercept formula:
y-1=2(x-(?/4)). I have followed numerous examples and I worked this problem exactly like the examples were worked, but I am not getting the right answer. The answer given is 4x-4y=?-4
For some reason, I am not following your equation.
y-1=2(x-(?/4)). I have followed numerous examples and I worked this problem exactly like the examples were worked, but I am not getting the right answer. The answer given is 4x-4y=?-4
For some reason, I am not following your equation.
D
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\(\displaystyle y \, = \, f'(a)\cdot (x - a) \, + \, f(a)\)
\(\displaystyle a = \frac{\pi}{4}\)
\(\displaystyle f'(a) = 2\)
\(\displaystyle f(a) \, = \, 1\)
\(\displaystyle y \, = \, 2\cdot (x - \frac{\pi}{4}) \, + \, 1\)
\(\displaystyle 2y \, - \, 2 \, = \, 4x \, - \, \pi\)
\(\displaystyle 2y \, - \, 4x \, = \, 2 \, - \, \pi\)
The answer in the book is apparently incorrect.
\(\displaystyle a = \frac{\pi}{4}\)
\(\displaystyle f'(a) = 2\)
\(\displaystyle f(a) \, = \, 1\)
\(\displaystyle y \, = \, 2\cdot (x - \frac{\pi}{4}) \, + \, 1\)
\(\displaystyle 2y \, - \, 2 \, = \, 4x \, - \, \pi\)
\(\displaystyle 2y \, - \, 4x \, = \, 2 \, - \, \pi\)
The answer in the book is apparently incorrect.