find the derivative of xcosy-ysinx+xytanx-xycoty=ycscy-xsecx

Question

please show steps, I missed school for this lesson today and am lost! thank you!

Answer 1

e are given the equation xcosy - ysinx +xytanx - xycoty = ycscy - xsecx

In order to find its derivative, we have to differentiate both sides with respect to x. We have to firstly break down each terms and find its derivative independently.

Lets differentiate the LEFT SIDE first:

Finding derivative of xcosy:

d/dx(xcosy) = cosy + x(-siny)y' = cosy - xsiny * y'

Finding derivative of -ysinx:

d/dx(-ysinx) = -y'sinx - ycosx

Finding derivative of xytanx using the product rule:

d/dx(xytanx) = xy'tanx + ytanx + xysec²x

Finding derivative of -xycoty using the product rule:

d/dx(-xycoty) = -[xy'coty + ycoty + xy(-csc²y)y']

= -xy'coty - ycoty + xycsc²y*y^2

So the derivative of the left side combined is:

cosy - xsiny*y' - y'sinx - ycosx + xy'tanx +ytanx +xysec²x - xy'coty - ycoty

Lets group the y' terms:

y'(-xsiny - sinx + xtanx -xcoty + xycsc²y) + cosy - ycosx + ytanx + xysec²x - ycoty

Now lets differentiate the RIGHT SIDE:

Finding derivative of ycscy:

d/dx(ycscy) = y'cscy + y(-cscycoty)y' = y'(cscy - ycscycoty)

Finding derivative of -xsecx:

d/dx(-xsecx) = -secx -xsecxtanx

So, the derivative of the right side combined is:

y'(cscy - ycscycoty) - secx - xsecxtanx

FINAL STEP:

We can equate the derivatives of both sides

y'(-xsiny - sinx + xtanx - xcoty + xycsc²y) + cosy - ycosx + ytanx + xysec²x - ycoty

= y'(cscy - ycscycoty) - secx - xsecxtanx

Now, lets move all y' terms to one side and everything else to the other:

y'[-xsiny - sinx + xtanx - xcoty + xycsc²y - cscy + ycscycoty]

= -secx - xsecxtanx - cosy + ycosx - ytanx - xysec²x + ycoty

So, the derivative dy/dx = y' is:

(-secx - xsecxtanx - cosy + ycosx - ytanx -xysec²x + ycoty) / (-xsiny - sinx + xtanx - xcoty + xycsc²y - cscy + ycscycoty)