Implicit differentiation problem

[Scorpy]

Hello everybody,

I've got this and one similar question in my exams, and I always get the wrong answer and zero points

y = 2 sin (pi x - y)

y' = 2 cos (pi x - y) (pi - y')

y'[1 + 2 cos (pi x - y)] = 2 pi cos (pi x - y)

=> y' = 2 pi cos (pi x - y) / [1 + 2 cos (pi x - y)

And now, the problem. The problem is the third line of this "code", because I don't know how the 1 appeared and everything else here
Can somebody help me with this? Explain me to understand it?

Thanks

 

[pka]

When we factor \(\displaystyle a+ab\) we get \(\displaystyle a(1+b)\).
Let \(\displaystyle a=y'\).

 

[Scorpy]

When we factor \(\displaystyle a+ab\) we get \(\displaystyle a(1+b)\).
Let \(\displaystyle a=y'\).

Okay, this is ok. But what about the term on the RHS? How can I get that? I do not understand

 

[soroban]

Hello, Scorpy!

\(\displaystyle y \:=\: 2\sin(\pi x - y)\)

\(\displaystyle y' \:=\: 2\cdot\cos(\pi x - y)\cdot (\pi - y')\)

We have: .. . . . . . .\(\displaystyle y' \:=\2\pi - 2y')\cdot\cos(\pi x - y)\)

. . . . . . . . . . . . . . \(\displaystyle y' \:=\:2\pi\cos(\pi - y) - 2y'\cos(\pi x - y)\)

.\(\displaystyle y' + 2y'\cos(\pi x - y) \:=\:2\pi\cos(\pi x - y)\)

\(\displaystyle y\big[1 + 2\cos(\pi x - y)\big] \:=\:2\pi\cos(\pi x - y)\)

. . . . . . . . . . . . . . \(\displaystyle y' \:=\:\dfrac{2\pi\cos(\pi x -y)}{1 + 2\cos(\pi x - y)}\)

 

[Scorpy]

Hello, Scorpy!


We have: .. . . . . . .\(\displaystyle y' \:=\2\pi - 2y')\cdot\cos(\pi x - y)\)

. . . . . . . . . . . . . . \(\displaystyle y' \:=\:2\pi\cos(\pi - y) - 2y'\cos(\pi x - y)\)

.\(\displaystyle y' + 2y'\cos(\pi x - y) \:=\:2\pi\cos(\pi x - y)\)

\(\displaystyle y\big[1 + 2\cos(\pi x - y)\big] \:=\:2\pi\cos(\pi x - y)\)

. . . . . . . . . . . . . . \(\displaystyle y' \:=\:\dfrac{2\pi\cos(\pi x -y)}{1 + 2\cos(\pi x - y)}\)

Thanks a lot!! I never try to multiply and rearrange the members of the equation My bad
And now, it's very simply. Thx again