Exploiting the Trifecta

Big Idea

Three powerful results in differentiation handle more exotic kinds of differentiation, the product rule:

$\displaystyle \frac{\mathrm d}{\mathrm dx}(uv) = v \cdot \frac{\mathrm du}{\mathrm dx} + u \cdot \frac{\mathrm dv}{\mathrm dx},$

the quotient rule:

$\displaystyle \frac{\mathrm d}{\mathrm dx} \left(\frac uv \right) = \frac{\displaystyle v \cdot \frac{\mathrm du}{\mathrm dx} - u \cdot \frac{\mathrm dv}{\mathrm dx}}{v^2},$

and the chain rule: if $y = f(u)$ , then

$\displaystyle \frac{\mathrm d}{\mathrm dx}(f(u)) = f'(u) \cdot \frac{\mathrm du}{\mathrm dx} \quad \iff \quad \frac{\mathrm dy}{\mathrm dx}=\frac{\mathrm dy}{\mathrm du} \cdot \frac{\mathrm du}{\mathrm dx}.$

Questions

Question 1. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} (2^{2x} \cdot \sqrt{16x} \cdot \ln(x^8))$ .

(Click for Solution)

Solution. We first simplify the function

$\begin{aligned} 2^{2x} \cdot \sqrt{16x} \cdot \ln(x^8) &= 4^x \cdot 4x^{1/2} \cdot 8 \ln(x) \\ &= 32 \cdot x^{1/2} \cdot 4^x \cdot \ln(x).\end{aligned}$

Apply the product rule once to obtain

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(x^{1/2} \cdot 4^x) &= 4^x \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) + x^{1/2} \cdot \frac{\mathrm d}{\mathrm dx}(4^x).\end{aligned}$

Apply the product rule on the original function to obtain

$\begin{aligned} & \frac{\mathrm d}{\mathrm dx} (2^{2x} \cdot \sqrt{16x} \cdot \ln(x^8)) \\ &= 32 \cdot \frac{\mathrm d}{\mathrm dx} ((x^{1/2} \cdot 4^x) \cdot \ln(x))\\ &= 32 \cdot \left( \ln(x) \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2} \cdot 4^x) + x^{1/2} \cdot 4^x \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \right) \\ &= 32 \cdot \left( \ln(x) \cdot \left( 4^x \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) + x^{1/2} \cdot \frac{\mathrm d}{\mathrm dx}(4^x) \right) + x^{1/2} \cdot 4^x \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \right) \\ &= 32 \cdot \left( 4^x \cdot \ln(x) \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) + x^{1/2} \cdot \ln(x) \cdot \frac{\mathrm d}{\mathrm dx}(4^x) + x^{1/2} \cdot 4^x \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \right) \\ &= 32 \cdot \left( 4^x \cdot \ln(x) \cdot \frac 12 x^{-1/2} + x^{1/2} \cdot \ln(x) \cdot 4^x \ln (4) + x^{1/2} \cdot 4^x \cdot \frac 1x \right) \\ &= 16 \cdot 4^x \cdot x^{-1/2} \cdot ( \ln(x) + 2x \cdot \ln(x) \cdot \ln (4) + 2 ).\end{aligned}$

Question 2. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} \left(\sin \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \right)$ .

(Click for Solution)

Solution. Noting that $\sin^{-1} \neq \csc = 1/{\sin}$ , applying the chain rule repeatedly and carefully,

$\begin{aligned} &\frac{\mathrm d}{\mathrm dx} \left(\sin \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \cdot \frac{\mathrm d}{\mathrm dx} \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \cdot \left( \frac{\mathrm d}{\mathrm dx} (\csc(x^2)) - \frac{\mathrm d}{\mathrm dx} (\sin^{-1}(x^2)) \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \cdot \left( -\csc(x^2) \cot(x^2) \cdot \frac{\mathrm d}{\mathrm dx}(x^2) - \frac 1{\sqrt{1 - (x^3)^2}} \cdot \frac{\mathrm d}{\mathrm dx}(x^3) \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} -\sin^{-1}(x^3) \right) \cdot \left( -\csc(x^2) \cot(x^2) \cdot 2x - \frac 1{\sqrt{1 - x^6}} \cdot 3x^2 \right) \\ &= \cos \left( \frac{1}{\sin(x^2)} - \sin^{-1}(x^3) \right) \cdot \left( -2x\csc(x^2) \cot(x^2) - \frac {3x^2}{\sqrt{1 - x^6}} \right). \end{aligned}$

Question 3. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}(x^x)$ .

(Click for Solution)

Solution. Using the laws of exponents, the chain rule, the product rule, linearity, and common derivatives,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}(x^x) &= \frac{\mathrm d}{\mathrm dx} (e^{x \ln(x)})\\ &= e^{x \ln(x)} \cdot \frac{\mathrm d}{\mathrm dx}( x \ln(x) ) \\ &= x^x \cdot \left( \frac{\mathrm d}{\mathrm dx}(x) \cdot \ln(x) + x \cdot \frac{\mathrm d}{\mathrm dx}(\ln(x)) \right) \\ &= x^x \cdot \left( 1 \cdot \ln(x) + x \cdot \frac{1}{x} \right) \\ &= x^x \cdot (\ln(x) + x). \end{aligned}$

—Joel Kindiak, 3 Sept 25, 1905H

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