Find the equation of the tangent line to the curve $\sin(xy) = x$ at the point $\left(1,\frac\pi2\right)$.

Question

Find the equation of the tangent line to the curve $\sin(xy) = x$ at the point $\left(1,\frac\pi2\right)$.

I was told I needed to differentiate with respect to $y$, and I want to see how that works and to solve the problem.

I tried to use implicit differentiation as usual, but the simplification gave me 1=0.

Answer 1

HINT

1. The equation of any line in $\mathbb{R}^2$ is given by $y - y_0 = m(x - x_0)$, where $(x_0,y_0)$ is any point on the line and $m$ is the slope of the line.
2. The slope of the tangent line to the curve $y = f(x)$ at a point $(x_0,y_0)$ is given by $f'(x_0)$.

So, compute $f'(x)$ and evaluate it at your point of interest, and then plug into the equation of the line and simplify... You may need implicit differentiation.

Answer 2

This curve is given by $\sin \left(xy\right) - x = 0$. Inspecting the gradient at this point, we have $\nabla f = \langle y\cos \left(xy\right) - 1, x\cos \left(xy\right) \rangle$. Plugging in to find the respective values of the slopes of $x, y$ we get $\langle \frac{π}{2}\cos \left(\frac{π}{2}*1\right) - 1, 1*\cos \left(\frac{π}{2}*1\right) \rangle = \langle -1, 0 \rangle$. Therefore the equation of the line is given by $-1\left(x - 1\right) + 0\left(y - \frac{π}{2}\right) = 0$ or $x = 1 $.

Answer 3

This curve is pretty peculiar

The yellow line is the tangent at $P\left(1;\;\dfrac{\pi}{2}\right)$

Its equation can be found using this formula

$${\left( {\frac{{\partial f}}{{\partial x}}} \right)_P}\left( {x - {x_P}} \right) + {\left( {\frac{{\partial f}}{{\partial y}}} \right)_P}\left( {y - {y_P}} \right) = 0$$ $f(x,y)=\sin (x y)-x;\;{\dfrac{{\partial f}}{{\partial x}}}=y \cos (x y)-1;\; {\dfrac{{\partial f}}{{\partial y}}}=x \cos (x y)$

Thus the tangent's equation is

$-(x-1)+0\cdot\left(y-\frac{\pi}{2}\right)=0\to x-1=0\to x=1$