How to find #dy/dx# from #x^2y+3xy^2-x=5#?

2 Answers
Mar 9, 2018

#dy/dx=(-2xy-3y^2+1)/(x^2+6xy)#

Explanation:

To solve this, we need to use implicit differentiation.

Since we are finding the derivative in respect to #x#, anytime we find the derivative of #y#, we need to tag on a #dy/dx# after.

To start off, we need to use the Product Rule for the first two terms, since there are two variables being multiplied together.

The Product Rule states:

#f'(x)g(x)+f(x)g'(x)#

So we can now set it up like this:

#x^2y+3xy^2-x=5#

#2xy+x^2dy/dx+3y^2+6xydy/dx-1=0#

Now that we have the derivative, we need to clean up the answer a bit. We need to move each term without the #dy/dx# to the other side. So we get:

#x^2dy/dx+6xydy/dx=-2xy-3y^2+1#

Now we factor out a #dy/dx#:

#dy/dx(x^2+6xy)=-2xy-3y^2+1#

Divide:

#dy/dx=(-2xy-3y^2+1)/(x^2+6xy)#

There is your answer

Mar 9, 2018

# dy/dx = (1 - 2xy - 3y^2)/(x^2 + 6xy) #

Explanation:

We have:

# x^2y+3xy^2-x=5 #

Differentiating wrt #x# by applying the product rule we get:

# x^2(d/dx y) + (d/dx x^2)y + (3x)(d/dx y^2) + (d/dx 3x)y^2 -d/dx x = d/dx 5 #

# :. x^2(d/dx y) + (2x)y + (3x)(d/dx y^2) + (3)y^2 -1 = 0 #

Then we apply the chain rule to perform the implicit differentiation:

# x^2(dy/dx) + (2x)y + (3x)(2y dy/dx) + (3)y^2 -1 = 0 #

Now, we collect terms and solve for #dy/dx#:

# (x^2 + 6xy)(dy/dx) = 1 - 2xy - 3y^2 #

# :. dy/dx = (1 - 2xy - 3y^2)/(x^2 + 6xy) #