# How do you find #dy/dx# by implicit differentiation given #y^2+3y-5=x#?

##### 1 Answer

# dy/dx = 1 /(2y+3)#

#### Explanation:

# y^2+3y−5=x #

Differentiate wrt

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

# :. d/dx(y^2) + 3dy/dx − 0 = 1 #

# :. d/dx(y^2) + 3dy/dx = 1 #

This is where we get stuck because we can differentiate

# dy/dx*d/dy(y^2) + 3dy/dx = 1 #

And we can now handle this:

# dy/dx*2y + 3dy/dx = 1 #

# :. 2ydy/dx + 3dy/dx = 1 #

# :. (2y+3)dy/dx = 1 #

# :. dy/dx = 1 /(2y+3)#

we can skip the direct application of the chain rule and use the result it yields which is that if we differentiate a function of