Differentiate both sidies of the equation: X = y' sin y' + cos y'

[newtagi]

Hi everyone. Could someone explain me how to differentiate both sides of the equation?

\(\displaystyle X\, =\, y'\, \sin\, y'\, +\, \cos\, y'\)

\(\displaystyle y'\, =\, \rho\)

\(\displaystyle X\, =\, \rho\, \sin\, \rho\, +\, \cos\, \rho\, \Rightarrow\, dx\, =\, d\left(\rho\, \sin\, \rho\, +\, \cos\,\rho\right)\)

\(\displaystyle dy\, =\, \rho\, dx\, \Rightarrow\, \int\, dy\, =\, \int\, \rho\, d\left(\rho\, \sin\, \rho\, +\, \cos\, \rho\right)\)

I know how to integrate and find derivative of any function, but I did not know how to differentiate. So, I wrote dx=d(psinp+cosp), then I did not know how to differentiate the right side of the equation. I mean, how can I possibly get rid of the brackets? Thank you in advance

 

[stapel]

Hi everyone. Could someone explain me how to differentiate both sides of the equation?

\(\displaystyle X\, =\, y'\, \sin\, y'\, +\, \cos\, y'\)

\(\displaystyle y'\, =\, \rho\)

\(\displaystyle X\, =\, \rho\, \sin\, \rho\, +\, \cos\, \rho\, \Rightarrow\, dx\, =\, d\left(\rho\, \sin\, \rho\, +\, \cos\,\rho\right)\)

\(\displaystyle dy\, =\, \rho\, dx\, \Rightarrow\, \int\, dy\, =\, \int\, \rho\, d\left(\rho\, \sin\, \rho\, +\, \cos\, \rho\right)\)

I know how to integrate and find derivative of any function, but I did not know how to differentiate.

"Finding the derivative" is "differentiating". What do you mean, specifically, when you say that you can do the one but not the other?

When you reply, please include the full and exact text of the original exercise, including the instructions. Thank you!

 

[HallsofIvy]

Differentiate with respect to what variable? y' means y differentiated with respect to what variable?

Also, while this is an equation and involves differentiation, it is not a "differential equation"! This should be in the "Calculus" forum.