How do you implicitly differentiate # e^(3x)=sin(x+2y) #?

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Answer 1 · 2016-04-04T11:16:42

#y'=(3e^(3x))/(2*cos(x+2y))-1/2#

From the given equation, just differentiate both sides of the equation with respect to #x#

#e^(3x)=sin(x+2y)#

#d/dx(e^(3x))=d/dx(sin(x+2y))#

#e^(3x)d/dx(3x)=cos(x+2y)*d/dx(x+2y)#

#(e^(3x))(3)=cos(x+2y)(1+2y')#

#(3e^(3x))/(cos(x+2y))=(cancelcos(x+2y)(1+2y'))/(cancelcos(x+2y))#

#(3e^(3x))/(cos(x+2y))=1+2y'#

#(3e^(3x))/(cos(x+2y))-1=2y'#

#((3e^(3x))/(cos(x+2y))-1)/2=(2y')/2#

#((3e^(3x))/(cos(x+2y))-1)/2=(cancel2y')/cancel2#

#y'=(3e^(3x))/(2*cos(x+2y))-1/2#

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