How do you implicitly differentiate #2=ytanx-xy #?

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Answer 1 · 2016-01-03T22:55:37

#dy/dx=(y-ysec^2x)/(tanx-x)#

Find the derivative of each part separately.

#d/dx(2)=0#

Next, use the product rule. Remember that when implicitly differentiating, the chain rule will kick in and spit out #y'# terms.

#d/dx(ytanx)=y'tanx+ysec^2x#

Again, use the product rule.

#d/dx(-xy)=-y-xy'#

Combine all the derivatives and solve for #y'#.

#0=y'tanx+ysec^2x-y-xy'#

#y-ysec^2x=y'(tanx-x)#

#y'=(y-ysec^2x)/(tanx-x)=dy/dx#