Q.Find $\dfrac{dy}{dx}$ if $y = \sqrt{4x^2 + 1}$.

A square root of a polynomial is a composite; write it as a power and use the chain rule.

With $u=4x^2+1$ ($u'=8x$): $\tfrac{dy}{dx}=\tfrac12(4x^2+1)^{-1/2}\cdot 8x=\dfrac{4x}{\sqrt{4x^2+1}}$.

Rewrite the root as a power: $y = (4x^2 + 1)^{1/2}$, a composite with inner $u = 4x^2 + 1$. Apply the chain rule (§3).

Differentiate the inner. $\dfrac{du}{dx} = 8x$.

Differentiate the outer power and multiply.

Check (dual-solve): verify numerically at $x = 1$. The formula gives $\dfrac{4(1)}{\sqrt{5}} \approx \dfrac{4}{2.23607} \approx 1.7889$. Directly, at $x = 1.001$, $\sqrt{4(1.002001)+1} = \sqrt{5.008004} \approx 2.237857$; at $x = 0.999$, $\sqrt{4.992004} \approx 2.234279$; slope $\approx \dfrac{2.237857 - 2.234279}{0.002} \approx 1.789$ — matching.

Errors: forgetting the inner derivative $8x$ (giving only $\tfrac{1}{2\sqrt{4x^2+1}}$); mishandling the fractional exponent (the derivative of $u^{1/2}$ is $\tfrac12 u^{-1/2}$, i.e. $\tfrac{1}{2\sqrt u}$); and trying to differentiate $\sqrt{4x^2+1}$ as $\sqrt{\tfrac{d}{dx}(4x^2+1)}$, which is meaningless.