Limit Definition of Derivative

Introduction

The derivative of the function  $f(x)$  is the instantaneous rate of change of  $f(x).$

Let  $h$ be a nonzero number.

The average rate of change of the function  $f(x)$  on the interval

Let's say we want to integrate

$\int x^{2}e^{x^{3}}~dx.$

Here, we can compute this antiderivative by using  $u-$substitution.

While  $u-$substitution is an important integration technique, it will not help us evaluate all integrals.

For example, consider the integral

$\int xe^{x}~dx.$

There is no substitution that will allow us to integrate this integral.

We need another integration technique called integration by parts.

The formula for integration by parts comes from the product rule for derivatives.

Recall from the product rule,

$(f(x)g(x))'=f'(x)g(x)+f(x)g'(x).$

Then, we have

${\begin{array}{rcl}\displaystyle {f(x)g(x)}&=&\displaystyle {\int f'(x)g(x)+f(x)g'(x)~dx}\\&&\\&=&\displaystyle {\int f'(x)g(x)~dx+\int f(x)g'(x)~dx.}\end{array}}$

If we solve the last equation for the second integral, we obtain

$\int f(x)g'(x)~dx=f(x)g(x)-\int f'(x)g(x)~dx.$

This formula is the formula for integration by parts.

But, as it is currently stated, it is long and hard to remember.

So, we make a substitution to obtain a nicer formula.

Let  $u=f(x)$  and  $dv=g'(x)~dx.$

Then,  $du=f'(x)~dx$  and  $v=g(x).$

Plugging these into our formula, we obtain

$\int u~dv=uv-\int v~du.$

Warm-Up

Evaluate the following integrals.

1)   $\int xe^{x}~dx$

Solution:
Using the Product Rule, we have
$f'(x)=(x^{2}+x+1)(x^{3}+2x^{2}+4)'+(x^{2}+x+1)'(x^{3}+2x^{2}+4).$
Then, using the Power Rule, we have
$f'(x)=(x^{2}+x+1)(3x^{2}+4x)+(2x+1)(x^{3}+2x^{2}+4).$
NOTE: It is not necessary to use the Product Rule to calculate the derivative of this function.
You can distribute the terms and then use the Power Rule.
In this case, we have
${\begin{array}{rcl}\displaystyle {f(x)}&=&\displaystyle {(x^{2}+x+1)(x^{3}+2x^{2}+4)}\\&&\\&=&\displaystyle {x^{2}(x^{3}+2x^{2}+4)+x(x^{3}+2x^{2}+4)+1(x^{3}+2x^{2}+4)}\\&&\\&=&\displaystyle {x^{5}+2x^{4}+4x^{2}+x^{4}+2x^{3}+4x+x^{3}+2x^{2}+4}\\&&\\&=&\displaystyle {x^{5}+3x^{4}+3x^{3}+6x^{2}+4x+4.}\end{array}}$
Now, using the Power Rule, we get
$f'(x)=5x^{4}+12x^{3}+9x^{2}+12x+4.$
In general, calculating derivatives in this way is tedious. It would be better to use the Product Rule.

Final Answer:
$f'(x)=(x^{2}+x+1)(3x^{2}+4x)+(2x+1)(x^{3}+2x^{2}+4)$
or equivalently
$f'(x)=x^{5}+3x^{4}+3x^{3}+6x^{2}+4x+4$

2)   $\int x\cos(2x)~dx$

Solution:
Using the Quotient Rule, we have
$f'(x)={\frac {x(x^{2}+x^{3})'-(x^{2}+x^{3})(x)'}{x^{2}}}.$
Then, using the Power Rule, we have
$f'(x)={\frac {x(2x+3x^{2})-(x^{2}+x^{3})(1)}{x^{2}}}.$
NOTE: It is not necessary to use the Quotient Rule to calculate the derivative of this function.
You can divide and then use the Power Rule.
In this case, we have
${\begin{array}{rcl}\displaystyle {f(x)}&=&\displaystyle {\frac {x^{2}+x^{3}}{x}}\\&&\\&=&\displaystyle {{\frac {x^{2}}{x}}+{\frac {x^{3}}{x}}}\\&&\\&=&\displaystyle {x+x^{2}.}\\\end{array}}$
Now, using the Power Rule, we get
$f'(x)=1+2x.$

Final Answer:
$f'(x)={\frac {x(2x+3x^{2})-(x^{2}+x^{3})}{x^{2}}}$
or equivalently
$f'(x)=1+2x$

3)   $\int \ln x~dx$

Solution:
Using the Quotient Rule, we get
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {\cos x(\sin x)'-\sin x(\cos x)'}{(\cos x)^{2}}}\\&&\\&=&\displaystyle {\frac {\cos x(\cos x)-\sin x(-\sin x)}{(\cos x)^{2}}}\\&&\\&=&\displaystyle {\frac {\cos ^{2}x+\sin ^{2}x}{\cos ^{2}x}}\\&&\\&=&\displaystyle {\frac {1}{\cos ^{2}x}}\\&&\\&=&\displaystyle {\sec ^{2}x}\end{array}}$
since  $\sin ^{2}x+\cos ^{2}x=1$  and  $\sec x={\frac {1}{\cos x}}.$
Since  ${\frac {\sin x}{\cos x}}=\tan x,$  we have
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \frac{d}{dx}{\tan x}=\sec^2 x.}

Final Answer:
Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\sec^2 x}

Exercise 1

Evaluate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x^3 e^{-2x}~dx.}

First, we need to know the derivative of  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \csc x.}   Recall

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \csc x =\frac{1}{\sin x}.}

Now, using the Quotient Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{\frac{d}{dx}(\csc x)} & = & \displaystyle{\frac{d}{dx}\bigg(\frac{1}{\sin x}\bigg)}\\ &&\\ & = & \displaystyle{\frac{\sin x (1)'-1(\sin x)'}{\sin^2 x}}\\ &&\\ & = & \displaystyle{\frac{\sin x (0)-\cos x}{\sin^2 x}}\\ &&\\ & = & \displaystyle{\frac{-\cos x}{\sin^2 x}} \\ &&\\ & = & \displaystyle{-\csc x \cot x.} \end{array}}

Using the Product Rule and Power Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{1}{x^2}(\csc x-4)'+\bigg(\frac{1}{x^2}\bigg)'(\csc x-4)}\\ &&\\ & = & \displaystyle{\frac{1}{x^2}(-\csc x \cot x+0)+(-2x^{-3})(\csc x-4)}\\ &&\\ & = & \displaystyle{\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.} \end{array}}

So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\frac{-\csc x \cot x}{x^2}+\frac{-2(\csc x-4)}{x^3}.}

Exercise 2

Evaluate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int e^{3x}\sin (2x)~dx.}

Notice that the function  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)}   is the product of three functions.

We start by grouping two of the functions together. So, we have  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g(x)=(2x\sin x)\sec x.}

Using the Product Rule, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{g'(x)} & = & \displaystyle{(2x\sin x)(\sec x)'+(2x\sin x)'\sec x}\\ &&\\ & = & \displaystyle{(2x\sin x)(\tan^2 x)+(2x\sin x)'\sec x.} \end{array}}

Now, we need to use the Product Rule again. So,

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{g'(x)} & = & \displaystyle{2x\sin x\tan^2 x+(2x(\sin x)'+(2x)'\sin x)\sec x}\\ &&\\ & = & \displaystyle{2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.} \end{array}}

So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle g'(x)=2x\sin x\tan^2 x+(2x\cos x+2\sin x)\sec x.}

But, there is another way to do this problem. Notice

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{g(x)} & = & \displaystyle{2x\sin x\sec x}\\ &&\\ & = & \displaystyle{2x\sin x\frac{1}{\cos x}}\\ &&\\ & = & \displaystyle{2x\tan x.} \end{array}}

Now, you would only need to use the Product Rule once instead of twice.

Exercise 3

Evaluate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x\sec^2 x~dx.}

Using the Quotient Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=\frac{(x^2\cos x+3)(x^2\sin x+1)'-(x^2\sin x+1)(x^2\cos x+3)'}{(x^2\cos x+3)^2}.}

Now, we need to use the Product Rule. So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{h'(x)} & = & \displaystyle{\frac{(x^2\cos x+3)(x^2(\sin x)'+(x^2)'\sin x)-(x^2\sin x+1)(x^2(\cos x)'+(x^2)'\cos x)}{(x^2\cos x+3)^2}}\\ &&\\ & = & \displaystyle{\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.} \end{array}}

So, we get

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle h'(x)=\frac{(x^2\cos x+3)(x^2\cos x+2x\sin x)-(x^2\sin x+1)(-x^2\sin x+2x\cos x)}{(x^2\cos x+3)^2}.}

Exercise 4

Evaluate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int x\sqrt{x+1}~dx.}

First, using the Quotient Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\ &&\\ & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.} \end{array}}

Now, we need to use the Product Rule. So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\ &&\\ & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.} \end{array}}

So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}

Exercise 5

Evaluate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \frac{\ln x}{x^3}~dx.}

First, using the Quotient Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\ &&\\ & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.} \end{array}}

Now, we need to use the Product Rule. So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\ &&\\ & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.} \end{array}}

So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}

Exercise 6

Evaluate  Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \int \sin(2x)\cos(3x)~dx.}

First, using the Quotient Rule, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x (e^x)'-e^x(x^2\sin x)'}{(x^2\sin x)^2}}\\ &&\\ & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\sin x)'}{x^4\sin^2 x}.} \end{array}}

Now, we need to use the Product Rule. So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{array}{rcl} \displaystyle{f'(x)} & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2(\sin x)'+(x^2)'\sin x)}{x^4\sin^2 x}}\\ &&\\ & = & \displaystyle{\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.} \end{array}}

So, we have

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f'(x)=\frac{x^2\sin x e^x - e^x(x^2\cos x+2x\sin x)}{x^4\sin^2 x}.}