Asked by Anonymous
Find
dy/dx
by implicit differentiation.
tan−1(4x2y) = x + 3xy2
dy/dx
by implicit differentiation.
tan−1(4x2y) = x + 3xy2
Answers
Answered by
Steve
Just use the product and chain rules:
arctan(4x^2y) = x + 3xy^2
1/(1+16x^4y^2) * (8xy+4x^2y') = 1 + 3y^2 + 6xyy'
8xy/(1+16x^4y^2) + 4x^2/(1+16x^4y^2) y' = 1+3y^2 + 6xyy'
y'(4x^2/(1+16x^4y^2) - 6xy) = 1+3y^2 - 8xy/(1+16x^4y^2)
At this point, it's just algebra, and serves no useful purpose to pursue it much further. Just solve for y' and you can massage it in various ways, as shown here:
http://www.wolframalpha.com/input/?i=derivative+arctan%284x^2y%29+%3D+x+%2B+3xy^2
arctan(4x^2y) = x + 3xy^2
1/(1+16x^4y^2) * (8xy+4x^2y') = 1 + 3y^2 + 6xyy'
8xy/(1+16x^4y^2) + 4x^2/(1+16x^4y^2) y' = 1+3y^2 + 6xyy'
y'(4x^2/(1+16x^4y^2) - 6xy) = 1+3y^2 - 8xy/(1+16x^4y^2)
At this point, it's just algebra, and serves no useful purpose to pursue it much further. Just solve for y' and you can massage it in various ways, as shown here:
http://www.wolframalpha.com/input/?i=derivative+arctan%284x^2y%29+%3D+x+%2B+3xy^2
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