Example 39: Differentiate $\sqrt{3x+2}+\frac{1}{\sqrt{2x^2+4}}$ and $\log_7(\log x)$

Basics you need

Part (i): $y = \sqrt{3x+2} + \frac{1}{\sqrt{2x^2+4}}$

1. Where is the function defined?

   The square root $\sqrt{3x+2}$ needs $3x + 2 \ge 0$⁠, that is $x \ge -\frac23$⁠. The second term needs $2x^2 + 4 > 0$ (positive, since it is under a root and in a denominator), and this always holds because $2x^2 \ge 0$ gives $2x^2+4 \ge 4$⁠. So

   $$y \text{ is defined for } x \ge -\tfrac23.$$
   The textbook says the function is defined for $x > -\frac23$⁠. Strictly it is also defined at $x = -\frac23$⁠, where $\sqrt{3x+2} = \sqrt0 = 0$⁠. What fails at $x = -\frac23$ is the derivative (Part (i), Step 5). So the book's condition $x > -\frac23$ is exactly the right condition for the answer, as it says at the end.
   A square root of a real number needs a non-negative input; a denominator must not be zero. The domain is the set of $x$ where every piece of the formula makes sense.

2. Rewrite both terms as powers

   $$y = (3x+2)^{1/2} + (2x^2+4)^{-1/2}.$$
   The power rule handles $u^n$ for any exponent, so rewriting roots and reciprocals as powers lets one rule do all the work. The minus sign in $-\frac12$ records "one over".
   $\sqrt u = u^{1/2}$ and $\frac{1}{u^{k}} = u^{-k}$ (laws of exponents).

3. Differentiate the first term (chain rule)

   Inner $t = 3x+2$⁠, $\frac{dt}{dx} = 3$⁠; outer $t^{1/2}$⁠, with derivative $\frac12 t^{-1/2}$⁠. For $x > -\frac23$ (so that $t > 0$⁠):

   $$\begin{aligned} \frac{d}{dx}(3x+2)^{1/2} &= \tfrac12(3x+2)^{-1/2}\cdot 3 \\ &= \frac{3}{2\sqrt{3x+2}}. \end{aligned}$$
   The power rule lowers the exponent by one, $\frac12 - 1 = -\frac12$⁠, and the chain rule multiplies by the inner rate 3. We need $t > 0$⁠, not just $t \ge 0$⁠, because $t^{-1/2} = \frac{1}{\sqrt t}$ is not defined at $t = 0$⁠.
   Power rule with $n = \frac12$⁠; chain rule, Theorem 4, Section 5.3.1. Lesson: the chain rule

4. Differentiate the second term (chain rule)

   Inner $s = 2x^2 + 4$⁠, $\frac{ds}{dx} = 4x$⁠; outer $s^{-1/2}$⁠, with derivative $-\frac12 s^{-3/2}$⁠. For every real $x$ (here $s \ge 4 > 0$⁠):

   $$\begin{aligned} \frac{d}{dx}(2x^2+4)^{-1/2} &= -\tfrac12(2x^2+4)^{-3/2}\cdot 4x \\ &= -\frac{2x}{(2x^2+4)^{3/2}}. \end{aligned}$$
   Lowering the exponent: $-\frac12 - 1 = -\frac32$⁠. Then $-\frac12\cdot 4x = -2x$⁠. The negative exponent $-\frac32$ moves to the denominator as $+\frac32$⁠. This term is differentiable everywhere because its inner function never reaches $0$⁠.
   Power rule with $n = -\frac12$⁠; $\frac{d}{dx}(2x^2+4) = 4x$ by the sum, constant-multiple and power rules (Section 5.3). Basics: derivative rules

5. Add the two derivatives, and check the end point

   By the sum rule, for $x > -\frac23$⁠:

   $$\frac{dy}{dx} = \frac{3}{2\sqrt{3x+2}} - \frac{2x}{(2x^2+4)^{3/2}}.$$

   At $x = -\frac23$ there is no derivative. Only a right-hand quotient is possible there (the function is not defined to the left), and for the first term it is

   $$\frac{\sqrt{3h} - 0}{h} = \frac{\sqrt3}{\sqrt h} \to \infty \quad (h \to 0^+).$$

   The second term has a finite derivative at $-\frac23$⁠, so the sum's quotient also grows without bound: the graph has a vertical tangent at its starting point.

   This is why the answer is stated for $x > -\frac23$ only. A formula with $\sqrt{3x+2}$ in a denominator could never be evaluated at $x = -\frac23$ anyway, and the difference quotient confirms that no finite slope exists there.
   Sum rule (1), Section 5.3; derivative as a limit of difference quotients, Section 5.3. Lesson: differentiability

Part (ii): $y = \log_7(\log x)$

1. Where is the function defined?

   The inner $\log x$ needs $x > 0$⁠. The outer $\log_7$ needs a positive input, so $\log x > 0$⁠, that is $x > e^0 = 1$⁠. So

   $$y \text{ is defined exactly for } x > 1.$$
   Logarithms of any base accept only positive numbers. Since $\log x$ is increasing and $\log 1 = 0$⁠, "$\log x > 0$" means "$x > 1$". This matches the textbook's statement.
   Domain of a logarithm: positive reals (Section 5.4). Basics: exponentials and logarithms

2. Change the base to $e$

   By the change of base rule, with $p = \log x$⁠:

   $$y = \log_7(\log x) = \frac{\log(\log x)}{\log 7}.$$
   Theorem 5 gives the derivative of the natural logarithm only. Changing base turns $\log_7$ into natural logarithms divided by the constant $\log 7 \approx 1.9459101$⁠. A constant divisor simply stays in front when we differentiate.
   Change of base (Section 5.4, property (1)): $\log_a p = \frac{\log_b p}{\log_b a}$⁠, here with $a = 7$ and $b = e$⁠. Basics: exponentials and logarithms

3. Differentiate $\log(\log x)$ by the chain rule

   Inner $s = \log x$⁠, $\frac{ds}{dx} = \frac1x$⁠; outer $\log s$⁠, with derivative $\frac1s$⁠. For $x > 1$ (so $s > 0$⁠):

   $$\frac{d}{dx}\log(\log x) = \frac{1}{\log x}\cdot\frac{1}{x}.$$
   Each link is a natural logarithm, whose derivative is "one over its input". The outer one sees the input $\log x$⁠, so its rate is $\frac{1}{\log x}$⁠, not $\frac1x$⁠.
   Theorem 5 (2), Section 5.4: $(\log x)' = \frac1x$⁠; chain rule, Theorem 4. Lesson: exponential and logarithmic functions

4. Put the constant back

   $$\begin{aligned} \frac{dy}{dx} &= \frac{1}{\log 7}\cdot\frac{1}{\log x}\cdot\frac1x \\ &= \frac{1}{x\log 7\,\log x}, \qquad x > 1. \end{aligned}$$
   A constant multiple $\frac{1}{\log 7}$ passes through the derivative unchanged. The final form multiplies the three denominators together. None of them is zero for $x > 1$⁠: $x > 0$⁠, $\log 7 > 0$⁠, $\log x > 0$⁠.
   Constant multiple rule, $(c\,u)' = c\,u'$⁠. Basics: derivative rules

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