Tedious Differentiation

Big Idea

The derivative of a function $f$ at a point $a$ , denoted $f'(c)$ , intuitively measures the gradient of the tangent to $y = f(x)$ at $(c, f(c))$ . For any general $x$ , we make the notation

$\displaystyle f'(x) \equiv \frac{\mathrm d}{\mathrm dx}(f(x)) \equiv \frac{\mathrm dy}{\mathrm dx}.$

For instance, given any real number $n$ ,

$\displaystyle \frac{\mathrm d}{\mathrm dx}(x^n) = nx^{n-1}.$

Several other derivatives of commonly used functions exist. Furthermore, differentiation is linear in the following sense:

$\displaystyle \frac{\mathrm d}{\mathrm dx}(af(x) + bg(x)) = a \cdot \frac{\mathrm d}{\mathrm dx}(f(x)) + b \cdot \frac{\mathrm d}{\mathrm dx}(g(x)).$

Questions

You may not use the chain rule in any of these problems.

Question 1. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx}\left(2 - 3 x + 4\sqrt x - \frac 5x\right)$ .

(Click for Solution)

Solution. Using the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx}\left(2 - 3 x + 4\sqrt x - \frac 5x\right) &= \frac{\mathrm d}{\mathrm dx}(2) - 3 \cdot \frac{\mathrm d}{\mathrm dx}(x) + 4 \cdot \frac{\mathrm d}{\mathrm dx}(\sqrt x) - 5 \cdot \frac{\mathrm d}{\mathrm dx} \left( \frac 1x \right) \\ &= \frac{\mathrm d}{\mathrm dx}(2) - 3 \cdot \frac{\mathrm d}{\mathrm dx}(x^1) + 4 \cdot \frac{\mathrm d}{\mathrm dx}(x^{1/2}) - 5 \cdot \frac{\mathrm d}{\mathrm dx} \left( x^{-1}\right) \\ &= 0 - 3 \cdot 1x^0 + 4 \cdot \frac 12 x^{-1/2} - 5 \cdot (-1)x^{-2} \\ &= -3 + 2 x^{-1/2} + 5x^{-2} \\ &= -3 + \frac 2{\sqrt x} + \frac 5{x^2}.\end{aligned}$

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

(Click for Solution)

Solution. Simplifying then using the linearity of differentiation,

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

Question 3. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} ((5+2x)^4)$ .

(Click for Solution)

Solution. We first slowly expand the function

$\begin{aligned} (5+2x)^4 &= ((5+2x)^2)^2 \\ &= (25 + 20x + 4x^2)^2 \\ &= 625 + 1000x + 600x^2 + 160x^3 + 16x^4. \end{aligned}$

Using the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} ((5+2x)^4) &= ((5+2x)^2)^2 \\ &= \frac{\mathrm d}{\mathrm dx} (625 + 1000x + 600x^2 + 160x^3 + 16x^4) \\ &= \frac{\mathrm d}{\mathrm dx}(625) + 1000 \cdot \frac{\mathrm d}{\mathrm dx}(x) + 600 \cdot \frac{\mathrm d}{\mathrm dx}( x^2 ) + 160 \cdot \frac{\mathrm d}{\mathrm dx}(x^3) + 16 \cdot \frac{\mathrm d}{\mathrm dx}( x^4) \\ &= 0 + 1000 \cdot 1 + 600 \cdot 2x + 160 \cdot 3x^2 + 16 \cdot 4x^3 \\ &= 1000 + 1200x + 480x^2 + 64x^3.\end{aligned}$

Question 4. Evaluate $\displaystyle \frac{\mathrm d}{\mathrm dx} \left( \left( \frac 2x-5 \sqrt x \right)^2 \right)$ .

(Click for Solution)

Solution. We first expand and simplify the function

$\begin{aligned} \left( \frac 2x-5 \sqrt x \right)^2 &= \left( \frac 2x \right)^2 - 2 \cdot \frac 2x \cdot 5\sqrt x + (5 \sqrt x)^2 \\ &= \frac{4}{x^2} - 20 \cdot x^{-1} \cdot x^{1/2} + 25 x \\ &= 4x^{-2} - 20 x^{-1/2} + 25 x. \end{aligned}$

Using the linearity of differentiation,

$\begin{aligned} \frac{\mathrm d}{\mathrm dx} \left( \left( \frac 2x-5 \sqrt x \right)^2 \right) &= \frac{\mathrm d}{\mathrm dx} (4x^{-2} - 20 x^{-1/2} + 25 x) \\ &= 4 \cdot \frac{\mathrm d}{\mathrm dx}(x^{-2}) - 20 \cdot \frac{\mathrm d}{\mathrm dx} (x^{-1/2}) + 25 \cdot \frac{\mathrm d}{\mathrm dx}(x) \\ &= 4 \cdot (-2)x^{-3} - 20 \cdot \left(-\frac 12\right)x^{-3/2} + 25 \cdot 1x^0 \\ &= -8x^{-3} + 10x^{-3/2}+25 \\ &= -\frac 8{x^3} + \frac{10}{\sqrt{x^3}}+25. \end{aligned}$

—Joel Kindiak, 3 Sept 25, 1826H

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