Q.Find $\dfrac{dy}{dx}$ if $y = \log(x^2 + 1)$.

Rewrite $\sqrt[3]{a^2+x^2}$ as $(a^2+x^2)^{1/3}$ and apply the chain rule; the answer is $\dfrac{2}{3}x(a^2+x^2)^{-2/3}$ — option (i).

Rewrite the cube root as a fractional power:

$y = \sqrt[3]{a^2 + x^2} = (a^2 + x^2)^{1/3}.$

Differentiate using the chain rule $\dfrac{d}{dx}[u^{1/3}] = \dfrac{1}{3}u^{-2/3}\cdot\dfrac{du}{dx}$ with $u = a^2 + x^2$, so $\dfrac{du}{dx} = 2x$:

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Applying the chain rule to $y = (\log x)^2$ gives $\dfrac{dy}{dx} = \dfrac{2\log x}{x}$.

The function is a composite: an outer function $u^2$ acting on the inner function $u = \log x$. The chain rule states $\dfrac{dy}{dx} = \dfrac{dy}{du}\cdot\dfrac{du}{dx}$.

Differentiate each part: …