Differentiate with respect to y : (2x-cos3y)^4sec(ln(1-xy))+√(1+(√y))?

1 Answer
Joel Kindiak · Becca M.
Aug 17, 2015

sec(ln(1−xy))(4(2x−cos3y)3(2dxdy+3sin3y))+(2x−cos3y)4⎛⎜⎝sec(ln(1−xy))tan(ln(1−xy))⋅(ydxdy+x)xy−1⎞⎟⎠+14√y+y√y

Explanation:

It's an atrocious chain rule phenomenon

Let (2x−cos3y)4sec(ln(1−xy))+√1+√y=pq+r
Where p=(2x−cos3y)4,q=sec(ln(1−xy)),r=√1+√y

dpdy=4(2x−cos3y)3(2dxdy+3sin3y)

dqdy=sec(ln(1−xy))tan(ln(1−xy))⋅⎛⎜⎝−(dxdyy+x)1−xy⎞⎟⎠

dqdy=sec(ln(1−xy))tan(ln(1−xy))⋅(ydxdy+x)xy−1

drdy=12√y2√1+√y=14√y+y√y

Thus ddy(2x−cos3y)4sec(ln(1−xy))+√1+√y
=ddy(pq+r)
=ddy(pq)+ddyr
=q(dpdy)+p(dqdy)+drdy
=sec(ln(1−xy))(4(2x−cos3y)3(2dxdy+3sin3y))+(2x−cos3y)4⎛⎜⎝sec(ln(1−xy))tan(ln(1−xy))⋅(ydxdy+x)xy−1⎞⎟⎠+14√y+y√y

Related questions
Impact of this question
1522 views around the world
You can reuse this answer
Creative Commons License