How do you implicitly differentiate $(xy)^2+cos(xy)=ln(xy)+tanh(xy)$ ?

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Answer 1 · 2016-05-02T10:29:03

$color(red)(y'=(1/x-2xy^2+y*sin(xy)+y*sech^2(xy))/(-1/y+2x^2y-x*sin(xy)-x*sech^2(xy)))$

We start from the given

$(xy)^2+cos(xy)=ln(xy)+tanh(xy)$

differentiate both sides of the equation

$d/dx(xy)^2+d/dxcos(xy)=d/dxln(xy)+d/dxtanh(xy)$

$2(xy)*(xy'+y*1)+(-sin(xy)*(xy'+y*1))=1/(xy)*(xy'+y*1)+sech^2(xy)*(xy'+y*1)$

$2x^2yy'+2xy^2-x*sin(xy)*y'-y*sin(xy)=(y')/y+1/x+x*sech^2(xy)*y'+y*sech^2(xy)$

$2x^2yy'-x*sin(xy)*y'-(y')/y-x*sech^2(xy)*y'=1/x+y*sech^2(xy)-2xy^2+y*sin(xy)$

$(2x^2y-x*sin(xy)-1/y-x*sech^2(xy))y'=1/x+y*sech^2(xy)-2xy^2+y*sin(xy)$

$color(red)(y'=(1/x-2xy^2+y*sin(xy)+y*sech^2(xy))/(-1/y+2x^2y-x*sin(xy)-x*sech^2(xy)))$

God bless....I hope the explanation is useful.

How do you implicitly differentiate (xy)^2+cos(xy)=ln(xy)+tanh(xy) (xy)^2+cos(xy)=ln(xy)+tanh(xy) ?