Question

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

Answer 1

`sec(ln(1-xy))(4(2x-cos3y)^3 (2 dx/dy+3 sin 3y)) + (2x-cos3y)^4(sec(ln(1-xy))tan(ln(1-xy))*((y dx/dy + x))/(xy-1)) + 1/(4sqrt(y+ysqrt(y)))`

Explanation:

It's an atrocious chain rule phenomenon

Let `(2x-cos3y)^4sec(ln(1-xy))+sqrt(1+sqrt(y))=pq+r`
Where `p=(2x-cos3y)^4, q=sec(ln(1-xy)), r=sqrt(1+sqrt(y))`

`(dp)/(dy) = 4(2x-cos3y)^3 (2 dx/dy+3 sin 3y)`

`(dq)/(dy) = sec(ln(1-xy))tan(ln(1-xy))*((-(dx/dy y + x))/(1-xy))`

`(dq)/(dy) = sec(ln(1-xy))tan(ln(1-xy))*((y dx/dy + x))/(xy-1)`

`(dr)/(dy) = (1/(2sqrt(y)))/(2sqrt(1+sqrt(y))) = 1/(4sqrt(y+ysqrt(y)))`

Thus `d/dy (2x-cos3y)^4sec(ln(1-xy))+sqrt(1+sqrt(y))`
`=d/dy (pq+r)`
`=d/dy (pq) + d/dy r`
`=q((dp)/(dy)) + p((dq)/(dy)) + (dr)/(dy)`
`=sec(ln(1-xy))(4(2x-cos3y)^3 (2 dx/dy+3 sin 3y)) + (2x-cos3y)^4(sec(ln(1-xy))tan(ln(1-xy))*((y dx/dy + x))/(xy-1)) + 1/(4sqrt(y+ysqrt(y)))`