First approach on derivatives
Average rate of change, difference quotient and instantaneous rate of change
Recall that the slope (or the rate of change) of the line connecting two points (x_0,y_0) and (x_1,y_1) is \dfrac{y_1-y_0}{x_1-x_0}.
Average rate of change
Let y=f\left(x\right) be a function. The average rate of change of f\left(x\right) over the interval [x_0,x_1] is:
\dfrac{f(x_1)-f(x_0)}{x_1-x_0}
The average rate of change of the function y=x^2 over the interval [0{,}3] is:
\dfrac{3^2-0^2}{3-0}=\dfrac{9}{3}=3
If y=mx+b is a linear function then the average rate of change of y over the interval [x_0,x_1] is:
\dfrac{mx_1+b-\left(mx_0+b\right)}{x_1-x_0}\\=\dfrac{m\left(x_1-x_0\right)}{x_1-x_0}\\=m
Therefore the average rate of change between any two points of a linear function is the slope of the linear function.
Difference quotient
Let y=f(x) be a function. Then the average rate of change of f(x) over the interval [x,x+h] is:
\dfrac{f\left(x+h\right)-f\left(x\right)}{x+h-x}
This particular average rate is called the difference quotient of the function:
\dfrac{f(x+h)-f(x)}{h}
The difference quotient of the function y=x^2 is:
\dfrac{(x+h)^2-x^2}{h}=\dfrac{x^2+2xh+h^2-x^2}{h}=\dfrac{2xh+h^2}{h}=2x+h
Let y=f(x) be a function and x_0 a real number. The smaller the value of h in the difference quotient \dfrac{f(x_0+h)-f(x_0)}{h}, the closer the difference quotient becomes to measuring the exact rate of change of f(x) at x_0.
If f(x)=x^2 then the average rate of change from 2 to 3 of f(x) is:
\dfrac{9-4}{3-2}=5
and the average rate of change from 2 to 4 of f(x) is:
\dfrac{16-4}{4-2}
Therefore the value 5 better measures the instantaneous rate of change of f(x) at x=2.
Instantaneous rate of change
The instantaneous rate of change of a function y=f(x) at x_0 is:
\lim\limits_{h\to 0}\dfrac{f(x_0+h)-f(x_0)}{h}
The average rate of change of f(x) from x_0 to x_1 measures the slope of the line which connects the points (x_0,f(x_0)) with (x_1,f(x_1)). The instantaneous rate of change of f(x) at x_0 measures the slope of the line at (x_0,f(x_0)) which is tangent to the graph of f(x).
The instantaneous rate of change of y=x^2 at 3 is:
\lim\limits_{h\to 0} \frac{(3+h)^2-3^2}{h}=\lim\limits_{h\to 0}2(3)+h=6
The instantaneous rate of change of y=\dfrac{1}{x} at x=2 is:
\lim\limits_{h\to 0}\dfrac{\frac{1}{2+h}-\frac{1}{2}}{h}=\lim\limits_{h\to 0}\dfrac{\frac{2-(2+h)}{(2+h)2}}{h}=\lim\limits_{h\to 0}\dfrac{-h}{(2+h)2h}=\lim\limits_{h\to 0}\dfrac{-1}{(2+h)2}=\dfrac{-1}{4}
Definition and notations
Derivative
The derivative of a function y=f(x) is the function \frac{df}{dx}=\lim\limits_{h\to 0} \frac{f(x+h)-f(x)}{h}, provided the limit exists and is not infinite. If the limit exists and is not infinite, then we say f(x) is differentiable.
The derivative of y=x^2 is:
\lim\limits_{h\to 0}\frac{(x+h)^2-x^2}{h}=\lim\limits_{h\to 0}2x+h=2x
A limit exists if and only if its left-hand and right-hand limits agree.
In the following example we show that the absolute value function is not differentiable at 0 by showing that a left-hand limit does not agree with a right-hand limit.
\lim\limits_{h\to 0^+}\frac{|x+h|-|x|}{h}=\lim\limits_{h\to 0^+}\frac{x+h-x}{h}=1
And;
\lim\limits_{h\to 0^-}\frac{|x+h|-|x|}{h}=\lim\limits_{h\to 0^-}\frac{-x-h-(-x)}{h}=-1
1\neq-1
The function f(x)=|x| is not differentiable at x=0.
We often write f'(x) instead of \dfrac{df}{dx} to denote the derivative of a function f(x). If a function is presented as y=f(x) we also let y' denote the derivative of f(x).
The instantaneous rate of change of the function y=\dfrac{1}{x} at x=7 is:
f'(7)=\dfrac{-1}{7^2}=\dfrac{-1}{49}
If y=\dfrac{1}{x} then:
y'=\lim\limits_{h\to 0}\dfrac{\frac{1}{x+h}-\frac{1}{x}}{h}=\lim\limits_{h\to 0}\dfrac{\frac{x-(x+h)}{(x+h)x}}{h}=\lim\limits_{h\to 0}\dfrac{-h}{(x+h)xh}=\lim\limits_{h\to 0}\frac{-1}{(x+h)x}=\frac{-1}{x^2}
Derivatives of usual functions
Derivative of constant or linear functions
Derivative of a constant function
The derivative of any constant function y=b (b\in\mathbb{R}) is:
y'=0
Consider the function f(x)=3. Then:
f'(x)=0
Derivative of a linear function
Let y=mx+b be a linear function. Then:
y'=m
If f(x)=7x-2 then:
f'(x)=7
The derivative of a linear function is the slope of its graph.
Derivative of a power function
Derivative of a power function
Let a be any real nonzero number. Then:
\dfrac{d}{dx}x^a=ax^{a-1}
The derivative of x^{2} is:
\dfrac{d}{dx} x^{2}=2 x
The derivative of x^{4} is:
\dfrac{d}{dx} x^{4}=4 x^{3}
The derivative of x^{1.7} is:
\dfrac{d}{dx} x^{1.7}=1.7 x^{0.7}
The derivative of \dfrac{1}{x} is:
\dfrac{d}{dx}\dfrac{1}{x}=\dfrac{d}{dx}x^{-1}=-1\cdot x^{-2}=\dfrac{-1}{x^2}.
The derivative of x^a is found by "bringing down the exponent" and then decreasing the exponent by 1.
Derivative of exponential and logarithmic functions
Derivative of exponential functions
Let a>0 and f(x)=a^x. Then :
\dfrac{df}{dx}= \ln(a) a^x
The derivative of e^x is:
\left(e^x\right)'=\ln(e)e^x=e^x
Therefore the derivative of e^x is itself.
The derivative of 2^x is:
\left(2^x\right)'=\ln (2)2^x.
The derivative of e^{2x} is:
\left(e^{2x}\right)'=\left((e^2)^x\right)'=\ln(e^2)e^{2x}=2e^{2x}.
Derivative of logarithmic functions
Let a>0 and f(x)=\log_a x. Then:
\dfrac{df}{dx}= \dfrac{1}{\ln (a) x}
The derivative of \ln x is:
\left(\ln x\right)'=\dfrac{1}{\ln(e)x}=\dfrac{1}{x}
The derivative of \log_{10}(x) is:
(\log_{10}(x))'=\dfrac{1}{\ln(10)(x)}
The derivative of \log_{e^2}x is:
(\log_{e^2}(x))'=\dfrac{1}{\ln(e^2)(x)}=\dfrac{1}{2x}.
Derivative of trigonometric functions
Derivative of \cos(x)
If f(x)=\cos(x) then:
\dfrac{df}{dx}=-\sin(x)
Derivative of \sin(x)
If f(x)=\sin(x) then:
\dfrac{df}{dx}=\cos(x)
When studying calculus of trigonometric functions, such as \sin(x) and \cos(x), it is important that we work in radians and not degrees. If we worked in degrees then the derivative of \sin(x) will not be \cos(x) and the derivative of \cos(x) will not be -\sin(x).
Higher-order derivatives
The derivative of the derivative of a function f(x) is called the second derivative of f(x) and is denoted as either \dfrac{d^2f}{dx^2}, f^{\left(2\right)}\left(x\right), or f''(x).
If f(x)=\sin(x) then:
f'\left(x\right)=\cos\left(x\right)
And:
f^{(2)}(x)=(\cos(x))'=-\sin(x)
If f(x)=mx+b is a linear function then:
f'\left(x\right)=m
And since f'(x) is a constant function:
f^{\left(2\right)}\left(x\right)=0
If n\in \mathbb{N} then the nth derivative of f(x) is denoted as either \dfrac{d^nf}{dx^n} or f^{(n)}(x).
The rules for finding the derivative of \sin(x) and \cos(x) allows us to compute the fourth derivative of \sin(x) as follows:
(\sin(x))^{(4)}=(\cos(x))^{(3)}=(-\sin(x))^{(2)}=(-\cos(x))^{(1)}=\sin(x)
If f(x)=a_nx^n+\cdots + a_1x+a_0 is a degree n polynomial then f'(x)=na_{n-1}x^{n-1}+\cdots + a_1+0 is a polynomial of degree n-1. Continuing to take derivatives we find that f^{(n+1)}(x)=0.
If f(x)=x^2-2x then:
f^{(3)}=(2x-2)''=(2)'=0
Rules of differentiation
Addition, subtraction and constant multiple of a function rules
Derivative of a function multiplied by a number
Let f(x) be a differentiable function and c a real number. Then:
(cf(x))'=cf'(x)
(20x^4)'=4\cdot 20x^3=80x^3
If f(x) is differentiable, then -f(x)=-1\cdot f(x) and therefore:
(-f(x))'=-f'(x)
The derivative of -\cos (x) is:
-1(\cos(x))'=-1(-\sin(x))=\sin(x)
Derivative of the addition of two functions
Let f(x) and g(x) be differentiable functions. Then f(x)+g(x) is differentiable and:
\left( f(x)+g(x)\right)'=f'(x)+g'(x)
The derivative of \sin(x)+x^2 is:
\((\sin(x)+x^2)'=(\sin(x))'+(x^2)'=\cos(x)+2x
Derivative of the substraction of two functions
Let f(x) and g(x) be differentiable functions. Then f(x)-g(x) is differentiable and:
\left( f(x)-g(x)\right)'=f'(x)-g'(x)
The derivative of \sin(x)-x^2 is:
\((\sin(x)-x^2)'=(\sin(x))'-(x^2)'=\cos(x)-2x
The same rules also apply when finding the derivative of the sums of three or more differentiable functions.
The derivative of 3x^2+7x^4-\sin(x) is:
\left(3x^2+7x^4-\sin(x)\right)'=(3x^2)'+(7x^4)'-(\sin(x))'=6x+28x^3-\cos(x)
Product and quotient rules
Product Rule
Let f(x) and g(x) be differentiable functions. Then the function f(x)g(x) is differentiable and:
(f(x)g(x))'=f'(x)g(x)+f(x)g'(x)
We want to determine f' when f\left(x\right)=\sin\left(x\right)\cos\left(x\right).
We first recall:
- (\sin(x))'=\cos(x)
- (\cos(x))'=-\sin(x).
Therefore:
(\sin(x)\cos(x))'=(\sin(x))'\cos(x)+\sin(x)(\cos(x))'=\cos^2(x)-\sin^2(x)
Quotient Rule
Let f(x) and g(x) be differentiable functions. Then the function \dfrac{f(x)}{g(x)} is differentiable at all values of x such that g(x)\not=0 and:
\left(\frac{f(x)}{g(x)}\right)'=\frac{f'(x)g(x)-f(x)g'(x)}{(g(x))^2}
We want to compute the derivative of f\left(x\right)=\dfrac{x^2+3x}{x+1}.
We first observe:
- ((x^2+3x)'=2x+3)
- (x+1)'=1
Therefore:
\left(\dfrac{x^2+3x}{x+1}\right)'=\dfrac{\left(x^2+3x\right)'\left(x+1\right)-\left(x^2+3x\right)\left(x+1\right)'}{\left(x+1\right)^2}=\dfrac{\left(2x+3\right)\left(x+1\right)-\left(x^2+3x\right)\left(1\right)}{\left(x+1\right)^2}=\dfrac{2x^2+5x+3-x^2-3x}{\left(x+1\right)^2}=\dfrac{x^2+2x+3}{\left(x+1\right)^2}
The quotient rule allows us to find the derivatives of the remaining trigonometric functions :
(\tan(x))'=\left(\frac{\sin(x)}{\cos(x)}\right)'=\frac{(\sin(x))'\cos(x)-\sin(x)(\cos(x))'}{\cos^2(x)}=\frac{\cos^2(x)+\sin^2(x)}{\cos^2(x)}=\frac{1}{\cos^2(x)}=\sec^2(x)
(\sec(x))'=\left(\frac{1}{\cos(x)}\right)'=\frac{(1)'\cos(x)-1(\cos(x))'}{\cos^2(x)}=\frac{\sin^2(x)}{\cos(x)}=\frac{\sin(x)}{\cos(x)}\frac{1}{\cos(x)}=\tan(x)\sec(x)
(\cot(x))'=\left(\frac{\cos(x)}{\sin(x)}\right)'=\frac{-\sin^2(x)-\cos^2(x)}{\sin^2(x)}=\frac{-1}{\sin^2(x)}=-\csc^2(x)
(\csc(x))'=\left(\frac{1}{\sin(x)}\right)'=\frac{0 \sin(x)-\cos(x)}{\sin^2(x)}=\frac{-\cos(x)}{\sin(x)}\frac{1}{\sin(x)}=-\cot(x)\csc(x)
Chain rule and inverse function rule
Chain rule
If f(x) and g(x) are differentiable functions, then so is the composition function f(g(x)) and:
(f(g(x)))'=f'(g(x))g'(x)
The function (x^3+1)^{100} is the composition of the functions x^{100} with x^3+1. Therefore, the chain rule allows us to find the derivative of (x^3+1)^{100} as follows:
((x^3+1)^{100})'=100(x^3+1)^{99}(x^3+1)'=100(x^3+1)^{99}(3x^2)=300(x^3+1)^{99}x^2
Inverse function rule
Suppose that f(x) is a differentiable function which has an inverse f^{-1}(x). If f'(x)\not=0 then f^{-1}(x) is differentiable and:
(f^{-1}(x))'=\dfrac{1}{f'(f^{-1}(x))}
Let f(x)=x^3+x^2+x.
- Observe that f(1)=3.
- We also observe that f'(x)=3x^2+2x+1 and f'(1)=6.
The inverse function rule will allow us to compute (f^{-1})'(3) as follows:
(f^{-1})'(3)=\frac{1}{f'(f^{-1}(3))}=\frac{1}{f'(1)}=\frac{1}{6}
If we restrict the domain of the function \sin(x) to only allow for x-values between -\pi and \pi, then \sin(x) becomes an invertible function and its inverse is denoted \arcsin(x). Observe that the domain of \arcsin(x) is [-1{,}1] while its range is [-\pi,\pi]. The inverse function rule allows us to find the derivative of \arcsin(x), as well as other inverse trigonometric functions.
Derivative of \arcsin(x)
Let f(x)=\arcsin(x), then:
f'(x)=\dfrac{1}{\sqrt{1-x^2}}
If f(x)=\arcsin(x^2) then:
f'(x)=\dfrac{1}{\sqrt{1-(x^2)^2}}\cdot (x^2)'=\dfrac{2x}{\sqrt{1-x^4}}
By using the inverse function rule and the fact that (\sin(x))'=\cos(x), we know that:
(\arcsin(x))'=\dfrac{1}{\cos(\arcsin(x))}
But \arcsin(x) is the unique angle \theta between -\pi and \pi such that:
\sin(\theta)=x
By the pythagorean identity:
\cos^2(\theta)+\sin^2(\theta)=\cos^2(\theta)+x^2=1
Therefore:
\cos(\theta)=\cos(\arcsin(x))=\sqrt{1-x^2}
Similar computations allow us to compute the derivates of the remaining inverse trigonometric functions:
(\arccos(x))'=\dfrac{-1}{\sqrt{1-x^2}}
(\arctan(x))'=\dfrac{1}{1+x^2}
(\mbox{arccot}(x))'=\dfrac{-1}{1+x^2}
(\mbox{arcsec}(x))'=\dfrac{1}{|x|\sqrt{x^2-1}}
(\mbox{arccsc}(x))'=\dfrac{-1}{|x|\sqrt{x^2-1}}
The derivative of \arctan(x^3) is:
(\arctan(x^3))'=\dfrac{1}{1+(x^3)^2}\cdot (x^3)'=\dfrac{3x^2}{1+x^6}
Implicit and logarithmic differentiation
Implicit differentiation
The chain rule provides other useful techniques for finding the derivatives of complicated functions. This process is referred to as implicit differentiation.
Implicit differentiation allows us to compute the rate of change of the unit circle. The unit circle satisfies the equation x^2+y^2=1 and implicit differentiation allows us to compute y':
(x^2+y^2)'=(1)'
2x+2yy'=0
2yy'=-2x
y'=\dfrac{-x}{y}
Suppose y is differentiable function satisfying the equation \sin(y)+x^2=\cos(x). Then by implicit differentiation:
(\sin(y)+x^2)'=\cos(y)y'+2x=(\cos(x))'=-sin(x)
Solving the equation for y' yields:
y'=\dfrac{-\sin(x)-2x}{\cos(y)}
Logarithmic differentiation
Suppose f(x) is a differentiable function. Then \ln(f(x)) is differentiable at all values of x such that f(x)>0 and by the chain rule and our rule for finding the derivative of logarithmic functions we have (\ln(f(x)))'=\dfrac{f'(x)}{f(x)}. Solving this equation for f'(x) provides us a rule for logarithmic differentiation.
Let f(x) be a differentiable function. Then:
f'(x)=f(x)(\ln(f(x)))'
Logarithmic differentiation is useful whenever we want to find the derivative of a function with a complicated exponent.
For example, to find the derivative of x^x we first find the derivative of \ln(x^x)=x\ln(x). We observe:
\(\displaystyle(x\ln(x))'=\ln(x)+x\dfrac{1}{x}=\ln(x)+1\)
By logarithmic differentiation we have:
(x^x)'=x^x(\ln(x^x))'=x^x(\ln(x)+1)=x^x\ln(x)+x^x
Review
| f(x) | f'(x) |
|---|---|
| a | 0 |
| mx+b | m |
| x^a | ax^{a-1} |
| a^x(a>0) | \ln(a)a^x |
| \log_a(x) | \dfrac{1}{\ln(a)x} |
| \cos(x) | -\sin(x) |
| \sin(x) | \cos(x) |
| \tan(x) | \sec^2(x) |
| \cot(x) | -\csc^2(x) |
| \sec(x) | \sec(x)\tan(x) |
| \csc(x) | -\csc(x)\tan(x) |
| \arcsin(x) | \dfrac{1}{\sqrt{1-x^2}} |
| \arccos(x) | \dfrac{-1}{\sqrt{1-x^2}} |
| \mbox{arctan}(x) | \dfrac{1}{1+x^2} |
| \mbox{arccot}(x) | \dfrac{-1}{1+x^2} |
| \mbox{arcsec}(x) | \dfrac{1}{|x|\sqrt{x^2-1}} |
| \mbox{arccsc}(x) | \dfrac{-1}{|x|\sqrt{x^2-1}} |
| ag(x)(a\in \mathbb{R}) | ag'(x) |
| g(x)+h(x) | g'(x)+h'(x) |
| g(x)h(x) | g'(x)h(x)+g(x)h'(x) |
| \dfrac{g(x)}{h(x)} | \dfrac{g'(x)h(x)-g(x)h'(x)}{(h(x))^2} |
| g(h(x)) | g'(h(x))h'(x) |