How do you find #(dy)/(dx)# given #sqrty+xy^2=5#?

1 Answer
Mar 15, 2018

#color(blue)(-(2y^(5/2))/(1+4xy^(3/2)))#

Explanation:

We need to differentiate this implicitly, because we don't have a function in terms of one variable.

When we differentiate #y# we use the chain rule:

#d/dy*dy/dx=d/dx#

As an example if we had:

#y^2#

This would be:

#d/dy(y^2)*dy/dx=2ydy/dx#

In this example we also need to use the product rule on the term #xy^2#

Writing #sqrt(y)# as #y^(1/2)#

#y^(1/2)+xy^2=5#

Differentiating:

#1/2y^(-1/2)*dy/dx+x*2ydy/dx+y^2=0#

#1/2y^(-1/2)*dy/dx+x*2ydy/dx=-y^2#

Factor out #dy/dx#:

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

Divide by #(1/2y^(-1/2)+2xy)#

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

Simplify:

Multiply by: #2sqrt(y)#

#(-y^2*2sqrt(y))/(2sqrt(y)1/(2sqrt(y))+2xy*2sqrt(y)#

#(-y^2*2sqrt(y))/(cancel(2sqrt(y))1/(cancel(2sqrt(y)))+2xy*2sqrt(y)#

#(-y^2*2sqrt(y))/(1+2xy*2sqrt(y))=-(2sqrt(y^5))/(1+4xsqrt(y^3))=color(blue)(-(2y^(5/2))/(1+4xy^(3/2)))#