Show tangent to y = x*tan(x) at x = pi/4 is (2 + pi)x - 2y = (pi)^2 / 4

[Axl_Adler]

So I've had a difficulty doing this H/W question for maths. This is the question:

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2. Show that the equation of the tangent to the curve \(\displaystyle y\, =\, x\, \tan(x)\) at \(\displaystyle x\, =\, \frac{\pi}{4}\) is

. . . . .\(\displaystyle \left(2\, +\, \pi\right)\, x\, -\, 2y\, =\, \dfrac{{\pi}^2}{4}\)

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So basically I assumed I should make y=mx+b equal to the equation given. To do that I derived xtanx and then plugged in the x value to find x. Then I solved y=mx+b, the issue is it was not equal to the other equation when I made y the subject. I know its hard to understand what I mean, the point is i tried and tried and can't get a answer. I'm not strong in maths and a lot of this is new to me so if someone could explain this step by step and also give me some resources to help with these types of questions it would be fantastic!

Thank you!

 

[stapel]

So I've had a difficulty doing this H/W question for maths. This is the question:

---

2. Show that the equation of the tangent to the curve \(\displaystyle y\, =\, x\, \tan(x)\) at \(\displaystyle x\, =\, \frac{\pi}{4}\) is

. . . . .\(\displaystyle \left(2\, +\, \pi\right)\, x\, -\, 2y\, =\, \dfrac{{\pi}^2}{4}\)

---

So basically I assumed I should make y=mx+b equal to the equation given. To do that I derived xtanx and then plugged in the x value to find x. Then I solved y=mx+b, the issue is it was not equal to the other equation when I made y the subject. I know its hard to understand what I mean...

Yes; it's very difficult to understand what you mean. For instance, what do you mean by "deriving" the given function, y = x*tan(x)? (How can you "obtain, via other steps based on other information" what you were already given? Or, by "derive", do you mean "differentiate"? And, if the latter, what did you get?)

...a lot of this is new to me so if someone could explain this step by step and also give me some resources to help with these types of questions it would be fantastic!

There are loads of great lessons online for finding the tangent line, via derivatives, to a given line at a point (here). The basic process is to find the actual coordinates of the point (by plugging the given x-value into the equation, and solving for the corresponding y-value). Then you find the slope at that point (by differentiating and then evaluating at the given x-value). Then you plug the slope you've found and the point you've found into any of the basic straight-line equations, and then rearrange to see if you can match their particular format for the line.

If you get stuck, please reply showing your work (rather than only making reference to it), so we can see where things are going sideways. Thank you!

 

[HallsofIvy]

I take it that, by "derived xtan x" you mean "found the derivative" or, more commonly, "differentiated" xtan(x). Yes, that's a good start! The tangent line to y= f(x), at \(\displaystyle x= x_0\), is \(\displaystyle y= f'(x_0)(x- x_0)+ f(x_0)\). What did you get for that derivative? What number did you get when you set x equal to \(\displaystyle \pi/4\)?

 

[Axl_Adler]

Thanks guys I solved it. It turns out I was putting the wrong thing in the calculator after I had found the derivative of y=xtanx.

Basically what I did was find the derivative, the find m. I plugged m into the y=mx+b formula and I substituted y for xtanx in the formula to find b.

At the end I substituted the x value and the y=mx+b equation into y (as the y value) into the given equation at the bottom. My explanation is confusing but at the end I got the answer, thank you!