3.1 Derivatives of Polynomials and Exponential Functions

In order to avoid using the limit definition of the derivative each time we want to differentiate a function, we develop some basic rules.

Power Functions

Theorem (Derivative of a Constant): Let \(c \in \mathbb{R}\). Then \(\frac{d}{dx} (c) = 0\).

Theorem (Derivative of a Power): Let \(n \in \mathbb{R}\) and \(f(x) = x^n\). Then \(f^{\prime}(x) = nx^{n-1}\).

Example: Let \(y = x^6\). Find \(y^{\prime}\).

Solution: Using the derivative of a power rule, \[y^{\prime} = 6x^{6-1} = 6x^5.\]

Theorem (Sum Rule): Let \(f(x)\) and \(g(x)\) be differentiable functions. Then \[\frac{d}{dx} \left[ f(x) \pm g(x)\right] = \frac{df}{dx} \pm \frac{dg}{dx}.\]

Example: Let \(y = x^4 + x^2 - x + 10\). Find \(y^{\prime}\).

Solution: Using the power rule and the sum rule, \[\begin{align*} y^{\prime} &= \frac{d}{dx} x^4 + x^2 - x + 10\\ &= 4x^{4-1} + 2x^{2-1} - 1x^{1-1} + 0\\ &= 4x^{3} + 2x - 1. \end{align*}\] Therefore, \(y^{\prime} = 4x^{3} + 2x - 1\).

Theorem (Constant Multiple Rule): Let \(f(x)\) be a differentiable function. Then \(\frac{d}{dx} cf(x)= c\frac{d}{dx} f(x)\).

Example: Let \(y = 5x^3 + 2x^{2.5} - 5x^{6/5} + 4 - 3x^{-1/3}\). Find \(\frac{dy}{dx}\).

Solution: Using our differentiation laws, \[\begin{align*} \dfrac{dy}{dx} &= \frac{d}{dx}5x^3 + 2x^{2.5} - 5x^{6/5} + 4 - 3x^{-1/3}\\ &= 5(3)x^{3-1} + 2(2.5)x^{2.5-1} - 5(6/5)x^{6/5-1} + 0 - 3(-1/3)x^{-1/3-1}\\ &= 15x^{2} + 5x^{1.5} - 6x^{1/5} + x^{-2/3}. \end{align*}\] Therefore, \(\frac{dy}{dx} = 15x^{2} + 5x^{1.5} - 6x^{1/5} + x^{-2/3}\).

Example: Let \(y = \frac{1}{3x^3} - \sqrt{x} + \sqrt[3]{x} + x^{\pi} - e^{\pi}\). Find \(\frac{dy}{dx}\).

Solution: Using our differentiation laws, \[\begin{align*} \dfrac{dy}{dx} &= \frac{d}{dx}\frac{1}{3x^3} - \sqrt{x} + \sqrt[3]{x} + x^{\pi} - e^{\pi}\\ &= \frac{d}{dx} 3x^{-3} -x^{1/2} + x^{1/3} + x^{\pi} - e^{\pi}\\ &= 3(-3)x^{-3-1} -(1/2)x^{1/2-1} + (1/3)x^{1/3-1} + \pi x^{\pi-1} - 0\\ &= -9x^{-4} -(1/2)x^{-1/2} + (1/3)x^{-2/3} + \pi x^{\pi-1}. \end{align*}\] Therefore, \(\frac{dy}{dx} = -9x^{-4} -(1/2)x^{-1/2} + (1/3)x^{-2/3} + \pi x^{\pi-1}\).

Practice Problems

  1. Find \(y^{\prime}\).
  • \(y = 4x^3 - 2x^2 + x - 10\)
  • \(y = 25x^2 - x^{-3} + x^{-5} + 5\)
  • \(y = x^{-1} + x^{-2/3} + x^{-4/3}\)
  • \(y = x^{4/5} - 3x^{10/7} + x^{3/5}\)
  1. Find \(\frac{dy}{dx}\).
  • \(y = \sqrt{x} + 4\sqrt[4]{x} - 3\sqrt[5]{x}\)
  • \(y = \dfrac{1}{\sqrt{x}} + 4\sqrt{x}\)
  • \(y = \dfrac{x^3 + x^2 + x + 1}{x}\)
  • \(y = (x^2 + 1)(x^3 - 1)\)

Proof of Differentiation Rules

Theorem (Derivative of a Constant): Let \(c \in \mathbb{R}\). Then \[\frac{d}{dx} (c) = 0.\]

Proof: Let \(c \in \mathbb{R}\) and \(f(x) = c\). Then \[\begin{align*} \frac{d}{dx} f(x) &= \lim_{h \rightarrow 0} \dfrac{f(x+h) - f(x)}{h}\\ &= \lim_{h \rightarrow 0} \dfrac{c - c}{h}\\ &= \lim_{h \rightarrow 0} \dfrac{0}{h}\\ &= \lim_{h \rightarrow 0} 0\\ &= 0. \end{align*}\] Therefore, \(\frac{d}{dx} (c) = 0\).

The power rule is true for all \(n \in \mathbb{R}\). At this point, we are unable to prove the general case. We can prove the power rule for \(n \in \mathbb{Z}^+\).

Theorem (Power Rule): Let \(n \in \mathbb{Z}^+\). Then \[\frac{d}{dx} x^n = nx^{n-1}.\]

Proof: Let \(n \in \mathbb{Z}^+\) and \(f(x) = x^n\). Then \[\begin{align*} \frac{d}{dx} f(x) &= \lim_{h \rightarrow 0} \; \dfrac{f(x+h) - f(x)}{h}\\ &= \lim_{h \rightarrow 0} \; \dfrac{(x+h)^n - x^n}{h}\\ &= \lim_{h \rightarrow 0} \; \dfrac{x^n + \frac{n(n-1)}{2}x^{n-1}h + \frac{n(n-1)(n-2)}{3!}x^{n-2}h^2 + \cdots + nxh^{n-1} + h^n- x^n}{h}\\ &= \lim_{h \rightarrow 0} \; \dfrac{nx^{n-1}h + \frac{n(n-1)}{2!}x^{n-2}h^2 + \cdots + nxh^{n-1} + h^{n}}{h}\\ &= \lim_{h \rightarrow 0} \; \dfrac{h\left[nx^{n-1} + \frac{n(n-1)}{2!}x^{n-2}h + \cdots + nxh^{n-2} + h^{n-1}\right]}{h}\\ &= \lim_{h \rightarrow 0} \; nx^{n-1} + \frac{n(n-1)}{2!}x^{n-2}h + \cdots + nxh^{n-2} + h^{n-1}\\ &= nx^{n-1}. \end{align*}\] Therefore, \(\frac{d}{dx} x^n = nx^{n-1}\).

Theorem (Sum Rule): Let \(f(x)\) and \(g(x)\) be differentiable functions. Then \[\frac{d}{dx} \left[ f(x) \pm g(x)\right] = \frac{df}{dx} \pm \frac{dg}{dx}.\]

Proof: Let \(f(x)\) and \(g(x)\) be differentiable and take \(k(x) = f(x) \pm g(x)\). Then \[\begin{align*} \frac{d}{dx} k(x) &= \lim_{h \rightarrow 0} \; \dfrac{k(x+h) - k(x)}{h}\\ &= \lim_{h \rightarrow 0} \; \dfrac{f(x+h) \pm g(x+h) - [f(x) \pm g(x)]}{h}\\ &= \lim_{h \rightarrow 0} \; \dfrac{[f(x+h) - f(x)] \pm [g(x+h) - g(x)]}{h}\\ &= \lim_{h \rightarrow 0} \; \dfrac{f(x+h) - f(x)}{h} \pm \dfrac{g(x+h) - g(x)}{h}\\ &= \lim_{h \rightarrow 0} \; \dfrac{f(x+h) - f(x)}{h} \pm \lim_{h \rightarrow 0} \; \dfrac{g(x+h) - g(x)}{h}\\ &= \dfrac{df}{dx} \pm \dfrac{dg}{dx}. \end{align*}\] Therefore, \(\frac{d}{dx} \left[ f(x) \pm g(x)\right] = \frac{df}{dx} \pm \frac{dg}{dx}\).

Theorem (Constant Multiple Rule): Let \(f(x)\) be a differentiable function. Then \[\frac{d}{dx} cf(x)= c\frac{d}{dx} f(x).\]

Proof: Let \(f(x)\) be differentiable and \(c \in \mathbb{R}\). Then \[\begin{align*} \frac{d}{dx} cf(x) &= \lim_{h \rightarrow 0} \; \dfrac{cf(x+h) - cf(x)}{h}\\ &= \lim_{h \rightarrow 0} \; \dfrac{c\left[f(x+h) - f(x)\right]}{h}\\ &= \lim_{h \rightarrow 0} \; c\dfrac{f(x+h) - f(x)}{h}\\ &= c \; \lim_{h \rightarrow 0} \; \dfrac{f(x+h) - f(x)}{h}\\ &= c \dfrac{d}{dx} f(x). \end{align*}\] Therefore, \(\frac{d}{dx} cf(x)= c\frac{d}{dx} f(x)\).

Derivatives of Exponential Functions

Theorem (Derivative of Exponential Functions): Let \(a \in \mathbb{R}\). Then \[\dfrac{d}{dx} a^x = a^x \ln(a) \;\;\; \text{ and } \;\;\; \dfrac{d}{dx} e^x = e^x.\]

We will prove this rule later in the book.

Example: Let \(y = 5e^x - 10^x\). Find \(\frac{dy}{dx}\).

Solution: Using our differentiation laws, \[\begin{align*} \dfrac{dy}{dx} &= \frac{d}{dx}\left(5e^x - 10^x\right)\\ &= 5\frac{d}{dx}e^x - \frac{d}{dx}10^x\\ &= 5e^x - 10^x \ln(10). \end{align*}\] Therefore, \(\frac{dy}{dx} = 5e^x - 10^x \ln(10)\).

Example: Let \(f(x) = 2^x + 2^{2x} + 2^{3x}\). Find \(f^{\prime}(x)\).

Solution: Using our differentiation laws, \[\begin{align*} f^{\prime}(x) &= \frac{d}{dx}\left(2^x + 2^{2x} + 2^{3x}\right)\\ &= \frac{d}{dx}2^x + \frac{d}{dx}2^{2x} + \frac{d}{dx}2^{3x}\\ &= \frac{d}{dx}2^x + \frac{d}{dx}4^{x} + \frac{d}{dx}8^{x}\\ &= 2^x\ln(2) + 4^{x}\ln(4) + 8^{x}\ln(8). \end{align*}\] Therefore, \(f^{\prime}(x) = 2^x\ln(2) + 4^{x}\ln(4) + 8^{x}\ln(8)\).

Practice Problems

  1. Find \(y^{\prime}\).
  • \(y = e^x + 4^x\)
  • \(y = 5 + 3^{3x}\)
  • \(y = 2^x + x^2 - 4\)
  • \(y = (e^x - e^x)(e^x + e^x)\)
  • \(y = e^x + x^e\)
  • \(y = 3e^x + \dfrac{2}{x} - \dfrac{5}{x^2}\)
  • \(y = \sqrt[4]{x} - 4e^x\)
  • \(y = e^{x+1} + 1\)
  • \(y = e^{2x}\)
  • \(y = e^{nx}\), for \(n \not = 0\).
  1. Find the equation of the tangent line at the indicated point. Sketch the graph and the tangent line.
  • \(y = e^x\) at \(a = 1\)
  • \(y = 2 + 3^{x}\) \(a = 0\)
  • \(y = 2^x - 4\) at \(a = 2\)
  • \(y = e^x + 2^x\) at \(a = 0\)