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If $y=(x^2+3)^{3/2}$, then $\dfrac{dy}{dx}$ is
A$\dfrac{3x}{2}(x^2+3)^{1/2}$
B$3x(x^2+3)^{1/2}$
C$\dfrac{3}{2}(x^2+3)^{3/2}$
D$3x(x^2+3)^{3/2}$
Answer & Solution
Correct answer: B. $3x(x^2+3)^{1/2}$
Apply the chain rule. If $y=(u)^{3/2}$ with $u=x^2+3$, then $\dfrac{dy}{dx}=\dfrac{3}{2}u^{1/2}\dfrac{du}{dx}=\dfrac{3}{2}(x^2+3)^{1/2}(2x)=3x(x^2+3)^{1/2}$.
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