# Differentiation of Implicit Function

Here you will learn what is the differentiation of implicit function with examples.

Let’s begin –

## Differentiation of Implicit Function

If the variables x and y are connected by a relation of the form f(x,y) = 0 and it is not possible or convenient to express y as a function x in the form y  = $$\phi (x)$$, then y is said to be an implicit function of x. To find $$dy\over dx$$ in such a case, we differentiate both sides of the given relation with respect to x, keeping in mind that the derivative of $$\phi (y)$$ with respect to x is $$d\phi\over dy$$.$$dy\over dx$$.

for example, $$d\over dx$$ (sin y) = cos y. $$dy\over dx$$ , $$d\over dx$$ ($$y^2$$) = 2y$$dy\over dx$$

It should be noted that $$d\over dy$$ (sin y) = cos y but $$d\over dx$$ (sin y) = cos y. $$dy\over dx$$.

Similarly , $$d\over dy$$ $$(y^3)$$ = $$3y^2$$ whereas $$d\over dx$$ $$(y^3)$$ = $$3y^2$$ $$dy\over dx$$.

Example : If $$ax^2 + 2hxy + by^2 + 2gx + 2fy + c$$ = 0, find $$dy\over dx$$.

Solution : We have,

$$ax^2 + 2hxy + by^2 + 2gx + 2fy + c$$ = 0

Differentiating both sides of this with respect to x, we get

$$d\over dx$$ $$(ax^2)$$ + $$d\over dx$$(2hxy) + $$d\over dx$$$$(by^2)$$ + $$d\over dx$$(2gx) + $$d\over dx$$(2fy) + $$d\over dx$$(c) = $$d\over dx$$(0)

$$\implies$$ a$$d\over dx$$$$(x^2)$$ + 2h$$d\over dx$$(xy) + b$$d\over dx$$$$(y^2)$$ + 2g$$d\over dx$$(x) + 2f$$d\over dx$$(y) + 0 = 0

$$\implies$$ 2ax + 2h$$(x{dy\over dx} + y)$$ + b 2y$$dy\over dx$$ + 2g.1 + 2f.$$dy\over dx$$ = 0

= $$dy\over dx$$(2hx + 2by + 2f) + 2ax + 2hy + 2g = 0

$$dy\over dx$$ = -$$2(ax + hy + g)\over 2(hx + by + f)$$

$$dy\over dx$$ = -$$(ax + hy + g)\over (hx + by + f)$$