Question

Implicit Differentiation.

x cos y = sin (x+y)

Answer 1

With respect to \(x\)?

Answer 2

Yes, dy/dx

Answer 3

I have cos y + x(-sin y) (y') = cos (x+y)(y') for the first line, but not sure if that is correct.

Answer 4

\[\begin{align*}x\cos y&=\sin(x+y)\\
\frac{d}{dx}[x\cos y]&=\frac{d}{dx}\sin(x+y)\\
\frac{d}{dx}[x]\cos y+x\frac{d}{dx}[\cos y]&=\cos(x+y)\frac{d}{dx}[x+y]\\
\cos y+x(-\sin y)\frac{dy}{dx}&=\cos(x+y)\left(\frac{d}{dx}[x]+\frac{dy}{dx}\right)\\
\cos y-x\sin y\frac{dy}{dx}&=\cos(x+y)\left(1+\frac{dy}{dx}\right)\\
\cos y-x\sin y\frac{dy}{dx}&=\cos(x+y)+\cos(x+y)\frac{dy}{dx}\\
\cos y-\cos(x+y)&=x\sin y\frac{dy}{dx}+\cos(x+y)\frac{dy}{dx}\\
\cos y-\cos(x+y)&=(x\sin y+\cos(x+y))\frac{dy}{dx}\\
\frac{\cos y-\cos(x+y)}{x\sin y+\cos(x+y)}&=\frac{dy}{dx}

\end{align*}\]

Answer 5

In case it's not clear, here are the details.

First line: given equation.
Second: differentiate both sides.
Third: product rule on left, chain rule on right.
Fourth: simplification and chain rule on left, distribution of differential operator on right.
Fifth: simplification on both sides.
Sixth: distribution of \(\cos(x+y)\) on right.
Seventh: exchange some terms.
Eighth: factor out \(\dfrac{dy}{dx}\) on right.
Ninth: isolate and solve for \(\dfrac{dy}{dx}\).

Answer 6

Oh, missed out on the chain rule on the right side. Thank you!

Answer 7

You're welcome!