# Find dy / dx, if x^2 + y^2 = sqrt 5.

## Question:

Find {eq}\dfrac {dy}{dx} {/eq}, if {eq}\displaystyle x^2 + y^2 = \sqrt 5 {/eq}.

## Differentiation in calculus:

In this problem, we need to find out the first derivative of the given function. The given function is an equation of circle. We'll solve this problem by using implicit differentiation and apply linearity.

Next, use the common derivative {eq}\dfrac{\ d}{ \ dx}(x^n)=nx^{n-1} {/eq}.

We are given:

{eq}x^2 + y^2 = \sqrt 5 {/eq}

Implicit differentiate both sides with respect to {eq}x {/eq}:

{eq}\displaystyle \Rightarrow \dfrac{\ d }{ \ dx } \left( x^2 + y^2 \right) = \dfrac{\ d }{ \ dx } ( \sqrt 5 ) {/eq}

Apply linearity:

{eq}\displaystyle \Rightarrow \dfrac{\ d }{ \ dx } \left(x^2 \right) + \dfrac{\ d }{ \ dx } ( y^2 ) = 0 {/eq}

Use the derivative power rule {eq}\displaystyle \dfrac{\ d }{ \ dx } (x^n)= n x^{n-1} {/eq}

{eq}\displaystyle \Rightarrow 2x + 2y\dfrac{\ dy }{ \ dx } = 0 {/eq}

{eq}\displaystyle \Rightarrow \dfrac{\ dy }{ \ dx }=- \dfrac{x}{y} {/eq}

Therefore the solution is: {eq}\displaystyle {\boxed{ \dfrac{\ dy }{ \ dx }=- \dfrac{x}{y} . }} {/eq}