Limit Definition of Derivative

Introduction

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

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int u~dv=uv-\int v~du.}

Warm-Up

Evaluate the following integrals.

1)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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)   Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle 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
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\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  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sin ^{2}x+\cos ^{2}x=1}   and  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \sec x={\frac {1}{\cos x}}.}
Since  ${\frac {\sin x}{\cos x}}=\tan x,$  we have
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\frac {d}{dx}}{\tan x}=\sec ^{2}x.}

Final Answer:
Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f'(x)=\sec ^{2}x}

Exercise 1

Evaluate  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \int x^{3}e^{-2x}~dx.}

First, we need to know the derivative of  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle \csc x.}   Recall

$\csc x={\frac {1}{\sin x}}.$

Now, using the Quotient Rule, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle f'(x)={\frac {-\csc x\cot x}{x^{2}}}+{\frac {-2(\csc x-4)}{x^{3}}}.}

Exercise 2

Evaluate  $\int e^{3x}\sin(2x)~dx.$

Notice that the function  $g(x)$  is the product of three functions.

We start by grouping two of the functions together. So, we have  Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle g(x)=(2x\sin x)\sec x.}

Using the Product Rule, we get

${\begin{array}{rcl}\displaystyle {g'(x)}&=&\displaystyle {(2x\sin x)(\sec x)'+(2x\sin x)'\sec x}\\&&\\&=&\displaystyle {(2x\sin x)(\tan ^{2}x)+(2x\sin x)'\sec x.}\end{array}}$

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

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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  $\int x\sec ^{2}x~dx.$

Using the Quotient Rule, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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  $\int x{\sqrt {x+1}}~dx.$

First, using the Quotient Rule, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 xe^{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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}(\sin x)'+(x^{2})'\sin x)}{x^{4}\sin ^{2}x}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.}\end{array}}}

So, we have

$f'(x)={\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.$

Exercise 5

Evaluate  $\int {\frac {\ln x}{x^{3}}}~dx.$

First, using the Quotient Rule, we have

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\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 xe^{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 (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}(\sin x)'+(x^{2})'\sin x)}{x^{4}\sin ^{2}x}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.}\end{array}}}

So, we have

$f'(x)={\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.$

Exercise 6

Evaluate  $\int \sin(2x)\cos(3x)~dx.$

First, using the Quotient Rule, we have

${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin x(e^{x})'-e^{x}(x^{2}\sin x)'}{(x^{2}\sin x)^{2}}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\sin x)'}{x^{4}\sin ^{2}x}}.}\end{array}}$

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

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle {\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}(\sin x)'+(x^{2})'\sin x)}{x^{4}\sin ^{2}x}}\\&&\\&=&\displaystyle {{\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.}\end{array}}}

So, we have

$f'(x)={\frac {x^{2}\sin xe^{x}-e^{x}(x^{2}\cos x+2x\sin x)}{x^{4}\sin ^{2}x}}.$