How do you implicitly differentiate # x^2+x/y-xy^2+x=3y #?

1 Answer
Jun 30, 2016

#(dy)/(dx)=(2xy^2+y-y^4+y^2 )/(3y^2+x+2xy^3) = f^'(x)#

Explanation:

#color(blue)("The teaching bit")#
By example: Suppose we had #y^2+3y=x^4#

Let #z=y^2+3y-> (dz)/(dy)= 2y+3#...................(1)
Then #z=x^4->(dz)/(dx)=4x^3#...............................(2)

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Consider equation (1). If we can convert #(dz)/(dy)# to #(dz)/(dx)# then we can directly equate the Eqn(1) to Eqn(2)

We use:#" "(dz)/(cancel(dy))xx(cancel(dy))/(dx)#

But from Eqn(1)

#(dz)/(dy)=2y+3" "# so #" "(dz)/(dy)xx(dy)/(dx)" "=" "(dy)/(dx)(2y+3)#

so we have #(dy)/(dx)(2y+2)=(dz)/(dx)=4x^3#

Thus #(dy)/(dx)=(4x^3)/(2y+2)#

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#color(blue)("Answering the question")#
Differentiating each term independently

As appropriate use #(d)/(dx)(uv)=v(du)/(dx)+u(dv)/(dx)#

As appropriate use #(d)/(dx)(u/v)=(v(du)/(dx)-u(dv)/(dx))/v^2#
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#d/(dx)(x^2)=color(brown)(+2x)#

#d/(dx)(x/y)=color(brown)( ( y-x(dy)/(dx))/y^2)#

#d/(dx)(-xy^2) = -[y^2+x2y(dy)/(dx)]=color(brown)(-y^2-2xy(dy)/(dx)#

#d/(dx)(x)=color(brown)(+1)#

#d/(dx)(3y) =color(brown)( 3(dy)/(dx))#
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#color(brown)("Putting it all together")#

The differentiated equation is:

#2x+(y-x(dy)/(dx))/y^2 -y^2-2xy(dy)/(dx) +1=3(dy)/(dx)#

Collecting like terms

#3(dy)/(dx) + (x(dy)/(dx))/y^2+2xy(dy)/(dx)=2x+cancel(y)/y^(cancel(2))-y^2+1 #

#(3y^2(dy)/(dx)+x(dy)/(dx)+2xy^3(dy)/(dx))/y^2=2x+1/y-y^2+1#

#(dy)/(dx)(3y^2+x+2xy^3)=y^2(2x+1/y-y^2+1)#

#(dy)/(dx)=(2xy^2+y-y^4+y^2 )/(3y^2+x+2xy^3)#