# Differentiation Rules

**Differentiation rules** are formulae that allow us to find the derivatives of functions quickly.

#### Contents

- Basic Properties of Derivatives
- Derivatives of Polynomials (Power Rule)
- Derivatives of Trigonometric Functions
- Derivatives of Exponential Functions
- Derivatives of Logarithmic Functions
- Chain Rule
- Product Rule
- Quotient Rule
- Inverse Functions
- Differentiation Rules Problem Solving - Basic
- Differentiation Rules Problem Solving - Intermediate
- Differentiation Rules - Problem Solving - Advanced

## Basic Properties of Derivatives

Taking derivatives of functions follows several basic rules:

- Multiplication by a constant: \( \big(c\cdot f(x)\big)' = c \cdot f'(x) \)
- Addition and subtraction: \( \big( f(x)\pm g(x)\big)' = f'(x) \pm g'(x) \)

For multiplication and composition of functions, see product rule and chain rule.

## Derivatives of Polynomials (Power Rule)

Main article: Power Rule

When taking the derivatives of polynomials, we can use the power rule:

Power Rule

\[ \frac{d}{dx} x^n = n\cdot x^{n-1} \]

## Derivatives of Trigonometric Functions

Main Article: Differentiation of Trigonometric Functions

We can see the basic trigonometric derivatives in the table below:

\( f(x) \quad \rightarrow \) | \( f'(x) \) |

\( \sin x \) | \( \cos x \) |

\( \cos x \) | \( -\sin x \) |

\( \tan x \) | \( \sec^2 x \) |

\( \sec x \) | \( \sec x \tan x \) |

\( \csc x \) | \( -\csc x \cot x \) |

\( \cot x \) | \( - \csc^2 x \) |

## Derivatives of Exponential Functions

Main Article: Differentiation of Exponential Functions

The main formula you have to remember here is the derivative of a logarithm:

\[\dfrac{d}{dx} \log_a x = \dfrac{1}{x \cdot \ln a}.\]

What is the derivative of the following exponential function:

\[ f(x) = 2 \cdot 3^x ?\]

We have

\[ \begin{align} f'(x) &= 2(3^x)' \\ &= 2\cdot 3^x \ln 3\\ &= 2(\ln 3)\cdot 3^x. \ _\square \end{align} \]

What is the derivative of the following exponential function:

\[ f(x) = 3e^x ?\]

We have

\[ \begin{align} f'(x) &= 3(e^x)' \\ &= 3e^x. \ _\square \end{align} \]

## Derivatives of Logarithmic Functions

Main Article: Differentiation of Logarithmic Functions

What is the derivative of the following logarithmic function:

\[ f(x) = \ln x + x ?\]

We have

\[ \begin{align} f'(x) &= (\ln x)' +x' \\ &= \frac{1}{x} + 1. \ _\square \end{align} \]

What is the derivative of the following logarithmic function:

\[ f(x) =3\log_{2} x -x^2 +2 ?\]

We have

\[ \begin{align} f'(x) &= (3\log_{2}x)'-\big(x^2\big)' +0 \\ &= 3\cdot \frac{1}{x} \cdot \frac{1}{\ln 2} - 2x \\ &= \frac{3}{x\ln 2} - 2x. \ _\square \end{align} \]

## Chain Rule

Main Article: Chain Rule

General form : \(\frac{d}{dx} f\big(g(x)\big) = f'\big(g(x)\big) \cdot g'(x)\)

What is the derivative of the following function:

\[ y=f(2x^2+x-3)?\]

We have

\[ \begin{align} y' &= f'(2x^2+x-3)\big(2x^2+x-3\big)' \\ &= (4x+1)f'(2x^2+x-3). \ _\square \end{align} \]

What is the derivative of the following function:

\[ f(x) = (x^2+x+1)^5 ?\]

We have

\[ \begin{align} f'(x) &= 5\big(x^2+x+1\big)^4\big(x^2+x+1\big)' \\ &= 5\big(x^2+x+1\big)^4(2x+1). \ _\square \end{align} \]

What is the derivative of the following function:

\[ y = \sqrt{2x+1} ?\]

Since \( y = \sqrt{2x+1} = (2x+1)^{\frac{1}{2}} ,\) we have

\[ \begin{align} y' &= \frac{1}{2}(2x+1)^{\frac{1}{2}-1} \cdot (2x+1)' \\ &= (2x+1)^{-\frac{1}{2}} \\ &= \frac{1}{\sqrt{2x+1}}. \ _\square \end{align} \]

## Product Rule

Main Article: Product Rule

What is the derivative of the following function:

\[ y = (x^2+3)(x^3+2x^2+5) ?\]

We have

\[ \begin{align} y' &= (x^2+3)'(x^3+2x^2+5) + (x^2+3)(x^3+2x^2+5)' \\ &= 2x(x^3+2x^2+5) + (x^2+3)(3x^2+4x) \\ &= 5x^4 + 8x^3 + 9x^2 +22x. \ _\square \end{align} \]

What is the derivative of the following function:

\[ y=(x^2+1)(2x+1)(3x+1) \]

when \(x = 1?\)

We have

\[ \begin{align} y' &= (x^2+1)'(2x+1)(3x+1) + (x^2+1)(2x+1)'(3x+1) + (x^2+1)(2x+1)(3x+1)' \\ &= 2x(2x+1)(3x+1) + (x^2+1)\cdot 2 \cdot (3x+1) + (x^2+1)(2x+1) \cdot 3 \\ &= 24x^3 + 15x^2 + 14x +5. \end{align} \]

Thus, the derivative of \(y\) at \(x=1\) is

\[y' \big\vert_{x=1} =24 \cdot 1^3 + 15 \cdot 1^2 + 14 \cdot 1 + 5 = 58. \ _\square\]

## Quotient Rule

Main Article: Quotient Rule

What is the derivative of the following function:

\[ y = \frac{1-x}{x^2+2} ?\]

We have

\[ \begin{align} y' &= \frac{(1-x)'(x^2+2)-(1-x)(x^2+2)'}{(x^2+2)^2} \\ &= \frac{(-1) \cdot (x^2+2)-(1-x) \cdot 2x}{(x^2+2)^2} \\ &= \frac{x^2-2x-2}{(x^2+2)^2}. \ _\square \end{align} \]

What is the derivative of the following function:

\[ y = \frac{1}{x^2\sqrt{x^3}} ?\]

Since \(y= \frac{1}{x^2\sqrt{x^3}} = x^{-\frac{7}{2}}, \) we have \[ \begin{align} y' &= -\frac{7}{2} x^{-\frac{9}{2}} \\ &= -\frac{7}{2x^4\sqrt{x}}. \ _\square \end{align} \]

## Inverse Functions

Main Article: Differentiation of Inverse Functions

If \(f(x) = x^3+ 7x+2,\) find \((f^{-1})'(10).\)

If \(f\) is a one-to-one differentiable function with inverse function \(f^{-1}\) and \(f'\big(f^{-1}(a)\big) \neq 0,\) then the inverse function is differentiable at \(a\) and

\[(f^{-1})'(a)=\frac{1}{f'\big(f^{-1}(a)\big)}.\]

Notice that \(f\) of this problem is one-to-one because

\[f'(x)=3x^2+7 > 0,\]

so \(f\) is increasing. To use the above theorem, we need to know \(f^{-1}(10)\) and we can find it by equating \(f(x)=10:\)

\[ \begin{align} 10 &= x^3 + 7x + 2 \\ \Rightarrow x &= 1, f^{-1}(10) = 1. \end{align} \]

Therefore,

\[(f^{-1})'(10) = \frac{1}{f'\big(f^{-1}(10)\big)} = \frac{1}{f'(1)} = \frac{1}{3\cdot 1^2 + 7} = \frac{1}{10}. \ _\square\]

## Differentiation Rules Problem Solving - Basic

If \( f(x+y) = f(x) + f(y) + xy \) and \(f'(0)=2,\) what is \(f'(2)?\)

Substituting \(x=0\) and \(y=0\) into \( f(x+y) = f(x) + f(y) + xy ,\) we get the value of \(f(0)\) as follows:

\[ f(0) = f(0) + f(0) + 0 \implies f(0) = 0 .\]

Thus,

\[ \begin{align} f'(2) &= \lim_{h\rightarrow 0} \frac{f(2+h)-f(2)}{h} \\ &= \lim_{h \rightarrow 0} \frac{f(2) + f(h) + 2h - f(2)}{h} \\ &= \lim_{h \rightarrow 0} \left ( \frac{f(h)}{h}+2\right ) \\ &= \lim_{h \rightarrow 0} \left ( \frac{f(0+h)-f(0)}{h}+2\right ) \\ &= f'(0) + 2 \\ &= 2+2 \\ &= 4. \ _\square \end{align} \]

## Differentiation Rules Problem Solving - Intermediate

If \(f(x) = \left(x + \sqrt{1+x^2}\right)^{10},\) what is \( f'(1)\cdot f'(-1) ?\)

We have

\[ \begin{align} f'(x) &= 10\left(x+\sqrt{1+x^2}\right)^9 \times \left(x+\sqrt{1+x^2}\right)' \\ &= 10\left(x+\sqrt{1+x^2}\right)^9\left(1 + \frac{2x}{2\sqrt{1+x^2}} \right ) \\ &= \frac{10}{\sqrt{1+x^2}}\left(x+\sqrt{1+x^2}\right)^{10}. \end{align} \]

Therefore,

\[ \begin{align} f'(1) \cdot f'(-1) &= \frac{10}{\sqrt{2}} \left(1+\sqrt{2}\right)^{10} \times \frac{10}{\sqrt{2}}\left(-1 + \sqrt{2}\right)^{10} \\ &= 50\left(\big(\sqrt{2} +1\big)\big(\sqrt{2} -1\big)\right)^{10} \\ &= 50(2-1)^{10} \\ &= 50. \ _\square \end{align} \]

Let \(f(x)\) and \(g(x)\) be two differentiable functions on \([0,2]\) such that

\[\begin{array}& f''(x) - g''(x) = 0, & f'(1) = 4, & f(2) = 9, & g'(1) = 2, & g(2) = 3. \end{array}\]

Find the value of \(f(x) - g(x)\) at \(x = \frac32.\)

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## Differentiation Rules - Problem Solving - Advanced

If \( F(t) = f\big(g(t)\big), f'(t) = \frac{1}{t^2+1}, g'(t) = \frac{10}{t^4 + 1},\) and \(g(0) = 3,\) what is \( F'(0) ?\)

Since \(F'(t) = f'\big(g(t)\big)g'(t),\)

\[ \begin{align} G'(0) &= f'\big(g(0)\big)g'(0) \\\\ &= f'(3)\cdot \frac{10}{0^4+1} \\\\ &= \frac{1}{3^2 +1}\cdot 10 \\\\ &= 1. \ _\square \end{align} \]

The function \(F(x)\) is defined by the following identity:

\[F\left(\big(F(x)+x\big)^k\right) = \big(F(x)+x\big)^2-x.\]

The value of \(F(1)\) is such that a finite number of possible numerical values of \(F'(1)\) can be determined solely from the above information. The maximum value of \(k\) such that \(F'(1)\) is an integer can be expressed as \(\frac{a}{b}\), where \(a\) and \(b\) are coprime integers.

What is the value of \(a+b?\)

What is the derivatives of following function:

\[ y= \sqrt[3]{(x+1)(x^2+1)} ,\] when \(x=0 ?\)

We have

\[ \begin{align} y &= \sqrt[3]{(x+1)(x^2+1)} \\ \Rightarrow y^3 &= (x+1)(x^2+1). \end{align} \]

Differentiating both sides with respect to \(x,\) we have

\[ \begin{align} 3y^2 \frac{dy}{dx} &= (x^2+1) +(x+1) \cdot 2x \\ &= 3x^2 +2x + 1 \\\\ \Rightarrow \frac{dy}{dx} &= \frac{3x^2 + 2x + 1}{3y^2} \\ &= \frac{3x^2+2x+1}{3 \sqrt[3]{(x+1)^2(x^2+1)^2}} \\\\ \Rightarrow \left. \frac{dy}{dx} \right\lvert_{x=0} &= \frac{1}{3}.\ _\square \end{align} \]

**Cite as:**Differentiation Rules.

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