curvesderivativestangent line
Take the graph $\sin(x) + \cos(y) =0.5$ (hint its really nasty)
I want to find the point where the tangent would be parrallel to the x-axis which I assume means where the gradient = 0
However I have no idea how to do the derivative of $\sin(x) + \cos(y) =0.5$
Any help would be great.
Thanks
One way is to recall the identity $\sin 2x=2\sin x\cos x$, which can be rewritten as $$\sin x\cos x=\frac{1}{2}\sin 2x.$$ Now differentiate, not forgetting to use the Chain Rule.
An alternative is to differentiate $\cos x\sin x$ immediately, using the Product Rule. We get $(\cos x)(\cos x)+(-\sin x)(\sin x)=\cos^2 x-\sin^2 x$. Then recall the identity $\cos 2x=\cos^2 x-\sin^2 x$.
Solving for gradient [i.e. "slope"] at $x = 0$ ($x = 0$ is that is the equation of the y-axis) given
$\;y^3+2y = \sin x+\cos x-1+3x^{2}\tag{1}$
Solve for $y$ at $x = 0$: $\quad x = 0 \implies y^3 + 2y = y(y^2 + 2) = 0\implies y = 0, \;x, y \in \mathbb R$
Use implicit differentiation on the left hand side and the right hand side: as a check, $$\frac{dy}{dx} = \frac{6x + \cos x - \sin x}{3y^2 + 2}$$
Evaluating the derivative at $(0, 0)$ will give the slope/gradient $m$ of the curve at the $y\text{-intercept}.\;$ As a check: we should get $m = 1/2.$
And here, in fact, is the graph of your function near the origin, and a graph of your function with the line tangent to the curve at $(0, 0)$
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Best Answer
Let $f(x,y)=\sin x+\cos y$. The tangent to a point of the curve $f(x,y)=\frac12$ will be horizontal when and only when the gradient of $f$ is vertivel, that is, when $\frac{\partial f}{\partial x}=0$. But $\frac{\partial f}{\partial x}(x,y)=\cos x$.