How do you differentiate #y = x * sqrt (4 - x^2)#?

Question

Cameron Alexander · Accepted Answer

# dy/dx =( 2 (2 - x^2 ))/sqrt(4 - x^2 ) #

Applying the "product rule" and "chain rule" to differentiate: #dy/dx = x . d/dx (4 - x^2)^(1/2 )+ (4 - x^2)^(1/2) .d/dx (x) # #dy/dx = x(1/2 (4 - x^2)^(-1/2) .d/dx (4 - x^2 ) )+ (4 - x^2 )^(1/2) . 1 # #dy/dx = x ( 1/2 (4 - x^2 )^(-1/2) .(-2x)) + (4 -x^2)^(1/2)# #dy/dx = - x^2 (4 - x^2)^(-1/2) + (4 - x^2 )^(1/2) # [ common factor of # (4 - x^2 )^(-1/2) ] # #dy/dx = (4 - x^2 )^(-1/2) [ -x^2 + 4 - x^2] # #dy/dx = (4 - x^2 )^(-1/2) . (4 - 2x^2) # #rArr dy/dx =( 2(2 - x^2))/sqrt(4 - x^2)#

Axel Alvarez · Answer

undefined To differentiate the function $ y = x \cdot \sqrt{4 - x^2} $, you can use the product rule and the chain rule.

First, let's denote $ u = x $ and $ v = \sqrt{4 - x^2} $.

Then, apply the product rule:

$$ \frac{dy}{dx} = u'v + uv' $$

Where $ u' $ is the derivative of $ u $ with respect to $ x $ and $ v' $ is the derivative of $ v $ with respect to $ x $.

Now, calculate the derivatives:

$$ u' = 1 $$
$$ v' = \frac{d}{dx}\left(\sqrt{4 - x^2}\right) = \frac{1}{2\sqrt{4 - x^2}} \cdot \frac{d}{dx}(4 - x^2) = \frac{-2x}{2\sqrt{4 - x^2}} = \frac{-x}{\sqrt{4 - x^2}} $$

Now, apply the product rule formula:

$$ \frac{dy}{dx} = x \cdot \frac{-x}{\sqrt{4 - x^2}} + \sqrt{4 - x^2} \cdot 1 $$

$$ \frac{dy}{dx} = \frac{-x^2}{\sqrt{4 - x^2}} + \sqrt{4 - x^2} $$

$$ \frac{dy}{dx} = \frac{-x^2 + 4 - x^2}{\sqrt{4 - x^2}} $$

$$ \frac{dy}{dx} = \frac{4 - 2x^2}{\sqrt{4 - x^2}} $$

So, the derivative of $ y = x \cdot \sqrt{4 - x^2} $ with respect to $ x $ is $ \frac{4 - 2x^2}{\sqrt{4 - x^2}} $.