Differentiating $\sin(x+y)+ \cos(x+y) = \log(x+y)$ with respect to $x$.

Question

It was given that $$\sin\left(x+y\right)+ \cos\left(x+y\right) = \log\left(x+y\right)$$ and I was asked to find $\frac{dy}{dx}$ for this. I attempted it as:

$1-$ $$\sin\left(x+y\right)+ \cos\left(x+y\right) = \log\left(x+y\right)$$ $2-$ $$\frac{d}{dx} \left(\sin\left(x+y\right)+ \cos\left(x+y\right)\right) =\frac {d}{dx} \log\left(x+y\right)$$ $3-$ $$\cos\left(x+y\right)\left(1+\frac{dy}{dx}\right) - \sin\left(x+y\right)\left(1+\frac{dy}{dx}\right) = \frac{ \left(1+\frac{dy}{dx}\right) }{x+y}$$

And now in step $4$, term $\left(1+\frac{dy}{dx}\right)$ gets cancelled from both sides and we are left with

$$\cos\left(x+y\right) - \sin\left(x+y\right)= \frac{1}{x+y}.$$

So we don't get any term containing $\frac{dy}{dx}$. So how can I find out $\frac{dy}{dx}$ if I am not getting it, same problem arises if I differentiate it again.

Answer 1

you can cancel $\left(1+\dfrac{dy}{dx}\right)$ only when $\left(1+\dfrac{dy}{dx}\right)\neq0$. But in this case: \begin{align*} \sin(x+y)+\cos(x+y)=\log(x+y)\\ \Rightarrow \dfrac{d}{dx}[\sin(x+y)]+\dfrac{d}{dx}[\cos(x+y)]&=\dfrac{d}{dx}[\log(x+y)]\\ \Rightarrow \cos(x+y)\cdot\left(1+\dfrac{dy}{dx}\right)-\sin(x+y)\cdot\left(1+\dfrac{dy}{dx}\right)&=\dfrac{1}{x+y}\cdot\left(1+\dfrac{dy}{dx}\right)\\ \Rightarrow\left(1+\dfrac{dy}{dx}\right)\left[\cos(x+y)-\sin(x+y)-\dfrac{1}{x+y}\right]&=0 \end{align*}

So either $\left(1+\dfrac{dy}{dx}\right)=0\quad$ or, $\quad\cos(x+y)-\sin(x+y)=\dfrac{1}{x+y}$.

Therefore, $\dfrac{dy}{dx}=-1\quad$ or, $\quad\cos(x+y)-\sin(x+y)=\dfrac{1}{x+y}$.

Answer 2

Notice that the equation you got from differentiating holds when $\frac{dy}{dx} = -1.$ This is your solution. Don't cancel that out, rearrange and cancel the other stuff out!

What's going on can be seen by briefly substituting $z$ in for $x+y$ and looking at $\cos(z)+\sin(z)=\log(z).$ You see this is a transcendental equation with maybe a couple solutions. So for any solution $z_0$ you have $y = z_0-x$ as a solution to your functional equation. Note that $dy/dx = -1.$ Solutions $y(x)$ generated this way are the only 'nice' solutions

(And it's the only solution provided there is only one solution to the transcendental equation. If there's a discrete set then you can do weird things like $y = z(x)-x$ where $z(x)$ hops between the solutions, but of course these solutions are not continuous.)

You notice when you cancel out the $(1+dy/dx)$ you have another transcendental equation that could hold instead, but that's unimportant since the solutions of interest all have $dy/dx=-1.$ (And furthermore it seems unlikely that $\cos(z)+\sin(z) = \log(z)$ and $\cos(z)-\sin(z) = 1/z$ have simultaneous solutions, so this second equation is likely irrelevant since it must simultaneously hold with the first. )