MHB What is the equation of the tangent line at (π/2,1) for sin(xy)=y?

Click For Summary
The discussion focuses on finding the equation of the tangent line at the point (π/2, 1) for the implicit function sin(xy) = y. To solve this, implicit differentiation is applied, treating y as a function of x, leading to the equation cos(xy)(y + xy') = y'. After isolating y', the expression obtained is y' = -cos(xy)y / (x cos(xy) - 1). Substituting the point (π/2, 1) results in y' = 0, confirming that the tangent line at this point is y = 1.
karush
Gold Member
MHB
Messages
3,240
Reaction score
5
Find the equation of the line tangent to
$$\sin\left({xy}\right)=y$$
At point
$$\left(\frac{\pi}{2 },1\right)$$
Answer $y=1$

I didn't know how to deal with xy.
No example given
 
Physics news on Phys.org
You need to differentiate implicitly, treating $y$ as $y(x)$. So, you'd start out with:
$$\d{}{x}\left[ \sin\left({xy}\right)=y \right] \qquad \implies \qquad \cos(xy) \cdot \left(y+xy'\right)=y'.$$
Can you continue from here?
 
Isolating $y'$ I got $$y '=\frac{-\cos\left({xy}\right)y}{x\cos\left({xy}\right)-1}$$
Plug in $$\left(\frac{\pi}{2 }, 1\right)\ \ y' =0$$
So $y=1$ is the equation
 
Last edited:
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

Similar threads

  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 1 ·
Replies
1
Views
1K
Replies
6
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
Replies
4
Views
2K