# How do you find the second derivative for the implicit equation #x^2+y^2 = a^2#?

##### 2 Answers

# (d^2y)/(dx^2) = -a^2/y^3 #

#### Explanation:

We have:

# x^2+y^2 = a^2 #

This represents a circle of radius

If we differentiate implicitly wrt

# 2x + 2ydy/dx = 0 #

# :. x + ydy/dx = 0 => dy/dx = -x/y#

Now, differentiating implicitly again, and applying the product rule, we get:

# :. 1 + (y)((d^2y)/(dx^2)) + (dy/dx)(dy/dx) = 0 #

# :. 1 + y(d^2y)/(dx^2) + (dy/dx)^2 = 0 #

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

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

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

# :. y(d^2y)/(dx^2) + a^2/y^2 = 0 #

# :. (d^2y)/(dx^2) = -a^2/y^3 #

#### Explanation:

Here, by the **Chain Rule,** we see that,

To get the

Here, for **Product Rule :**

But,

**Alternatively,** the same result can be obtained by diff.ing, w.r.t.

**Enjoy Maths.!**