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Worked solution

Differentiate a square-root quotient

Find dydx\dfrac{dy}{dx} if y=32x+1y=\sqrt{\dfrac3{2x+1}}.

[2007/SRJC/I/5]

CJC · Discussion · Question 16

Answer

dydx=−3(2x+1)3/2\dfrac{dy}{dx}=-\dfrac{\sqrt3}{(2x+1)^{3/2}}, for x>−1/2x>-1/2.

Rewrite as a power

y=3(2x+1)−1/2y=\sqrt3(2x+1)^{-1/2}, with real domain x>−1/2x>-1/2.

How did this part go?

Apply the chain rule

dydx=3(−1/2)(2x+1)−3/2(2)=−3(2x+1)3/2\dfrac{dy}{dx}=\sqrt3(-1/2)(2x+1)^{-3/2}(2)=-\dfrac{\sqrt3}{(2x+1)^{3/2}}.

How did this part go?

Common mistakes

The constant is √3, and the inner derivative contributes a factor of 2.

Read the hint