Question

Calculate dy/dx if 2cos^2 *y=xy

Answer 1

Calculate dy/dx if \[2\cos^2y=xy\]

Answer 2

differentiate each side
(note that the derivative of any "y" will have a chain rule of y' being that y is also a function of x)

Answer 3

What would be the derivative of \(\cos^2x\) ?
Once you calculate that, the derivative of \(\cos^2y\) will be the same, EXCEPT that it will be multiplied times y'.

Answer 4

so it would be
-2cos(x)sin(x)=y
?

Answer 5

\(\large\color{black}{ \displaystyle 2\cos^2x }\)
would have a derivative of
\(\large\color{black}{ \displaystyle -2\cos(x)\sin(x) }\)

AND,
\(\large\color{black}{ \displaystyle 2\cos^2y }\)
would have a derivative of
\(\large\color{black}{ \displaystyle -2\cos(y)\sin(y) \times y'}\)

Answer 6

this y', is simply a chain rule.

Answer 7

then the right side, do you know how to differentiate that?

Answer 8

It would end up either x^0y=y*1 or dy/dx[xy] so that'd be dy/d[y]? or something like that

Answer 9

the derivative of xy goes by the product rule:

\(\large\color{black}{ \displaystyle \frac{dy}{dx} \left[ xy \right]=x\frac{dy}{dx} \left[ y \right]+y\frac{dy}{dx} \left[ x \right] =xy'+y\cdot1}\)

Answer 10

So what would this derivative look like in final form? not too much simplification would be required, luckily.

Answer 11

yes, just:

\(\large\color{black}{ \displaystyle 2\cos^2 y=xy }\)

differentiate:

\(\large\color{black}{ \displaystyle -2y'\sin y\cos y=xy'+y }\)

solve for y'

\(\large\color{black}{ \displaystyle -2y'\sin y\cos y-xy'=y }\)

\(\large\color{black}{ \displaystyle y'(-2\sin y\cos y-x)=y }\)

\(\large\color{black}{ \displaystyle y'=y/(-2\sin y\cos y-x) }\)

Answer 12

\[\large\color{black}{ \displaystyle -\color{red}{4}y'\sin y\cos y=xy'+y }\]

Answer 13

oh, yeah, there is another 2, thanks for correcting me

Answer 14

\(\large\color{black}{ \displaystyle 2\cos^2 y=xy }\)

differentiate:

\(\large\color{black}{ \displaystyle -4y'\sin y\cos y=xy'+y }\)

solve for y'

\(\large\color{black}{ \displaystyle -4y'\sin y\cos y-xy'=y }\)

\(\large\color{black}{ \displaystyle y'(-4\sin y\cos y-x)=y }\)

\(\large\color{black}{ \displaystyle y'=y/(-4\sin y\cos y-x) }\)

Answer 15

Thanks, that really helped me out!