Differentiate w.r.t. $x$, the following function: (i) $\sqrt{3 x+2}+\frac{1}{\sqrt{2 x^{2}+4}}$ (ii) $\log _{7}(\log x)$

Step 1: Differentiate the first term of (i)

To differentiate the first term, $\sqrt{3x+2}$, we rewrite it as $(3x+2)^{1/2}$. We then apply the chain rule: first differentiate the outer power function, and then multiply by the derivative of the inner function $(3x+2)$.

Step 2: Differentiate the second term of (i)

For the second term, $\frac{1}{\sqrt{2x^2+4}}$, we rewrite it as $(2x^2+4)^{-1/2}$. Again, we use the chain rule: differentiate the outer power function, and then multiply by the derivative of the inner function $(2x^2+4)$.

Step 3: Combine results for (i)

The derivative of the sum of two functions is the sum of their derivatives. We combine the results from the previous two steps to get the final derivative for part (i).

Step 4: Differentiate (ii) using change of base and chain rule

For $\log_7(\log x)$, we first use the change of base formula for logarithms: $\log_b a = \frac{\log a}{\log b}$. Then, we apply the chain rule twice. First, differentiate with respect to $\log x$, and then differentiate $\log x$ with respect to $x$. Remember that $\frac{d}{dx}(\log x) = \frac{1}{x}$.

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