How do you implicitly differentiate #7xy- 3 lny= 42#?

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Answer 1 · 2017-02-18T02:06:52

#(dy/dx)=(-7y)/((7x) -(3/y))#

Ignore the #-3ln(y)# for now. Take the derivative of 7xy with product rule.

#(7x)(dy/dx)+(y)(7)#

#(7x)(dy/dx)+7y#

Now take a look at -3lny. Since 3 is a constant in front, you only need to worry about the derivaitve of lny and then multiply the 3 later.

#ln(y)=(1/y)(dy/dx)#

The derivative of a constant is always 0 so the equation will be equal to 0. Now combine your two parts together.

#(7x)(dy/dx)+7y -(3/y)(dy/dx)=0#

Subtract 7y on both sides.

#(7x)(dy/dx) -(3/y)(dy/dx)=-7y#

Then factor out #dy/dx# on the left side.

#(dy/dx)((7x) -(3/y))=-7y#

Isolate #dy/dx#.

#(dy/dx)=(-7y)/((7x) -(3/y))#