1 answer

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

Question:

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

Answers

#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)))#

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