Question

What is the slope of the tangent line of `sinx-sin(y^2)/x= C`, where C is an arbitrary constant, at `(pi/3,pi/2)`?

Answer 1

`(pi^2+18sin(pi^2/4))/(6pi^2sin(pi^2/4))`

Explanation:

The constant `C`
If we know that `(pi/3,pi/2)` is a point on the graph, then `C = sin(pi/3)-sin((pi/2)^2)/(pi/3) = sqrt3/2-(3sin((pi/2)^2))/pi`.
So `C` is determined by the values of `x` and `y`. It is not arbitrary. (It is not independent of the values of the variables.)

Slope of Tangent
Regardless of that, we can find the slope of the tangent line by implicit differentiation.

`d/dx(sinx-sin(y^2)/x)=d/dx( C)`

`cosx-((sin(y^2)*2y dy/dx)(x)-(sin(y^2))(1))/x^2 = 0`

`x^2cosx-2xysin(y^2)* dy/dx + sin(y^2) = 0`

`dy/dx = (x^2cosx+sin(y^2))/(2xysin(y^2))`

The slope at `(pi/3,pi/2)`, is therefore

`dy/dx = ((pi/3)^2cos(pi/3)+sin((pi/2)^2))/(2(pi/3)(pi/2)sin((pi/2)^2))`

`= (pi^2+18sin(pi^2/4))/(6pi^2sin(pi^2/4))`