Limit Definition of Derivative

Introduction

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

Let  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} be a nonzero number.

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

Let's say we want to integrate

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^2e^{x^3}~dx.}

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

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

For example, consider the integral

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 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,

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

Then, 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)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

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 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  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 u=f(x)}   and  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 dv=g'(x)~dx.}

Then,  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 du=f'(x)~dx}   and  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 v=g(x).}

Plugging these into our formula, we obtain

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

Warm-Up

Evaluate the following integrals.

1)   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 xe^x~dx}

Solution:
Using the Product 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 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
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)=(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
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{(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
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)=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:
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)=(x^2+x+1)(3x^2+4x)+(2x+1)(x^3+2x^2+4)}
or equivalently
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)=x^5+3x^4+3x^3+6x^2+4x+4}

2)   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\cos (2x)~dx}

Solution:
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 f'(x)=\frac{x(x^2+x^3)'-(x^2+x^3)(x)'}{x^2}.}
Then, using the 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 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
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+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 (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)=1+2x.}

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)=\frac{x(2x+3x^2)-(x^2+x^3)}{x^2}}
or equivalently
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)=1+2x}

3)   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 \ln x~dx}

Solution:
Using the Quotient 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{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 (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 \sin^2 x+\cos^2 x=1}   and  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 \sec x=\frac{1}{\cos x}.}
Since  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{\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  $\int x\sec ^{2}x~dx.$

Using the Quotient Rule, we have

$h'(x)={\frac {(x^{2}\cos x+3)(x^{2}\sin x+1)'-(x^{2}\sin x+1)(x^{2}\cos x+3)'}{(x^{2}\cos x+3)^{2}}}.$

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

${\begin{array}{rcl}\displaystyle {h'(x)}&=&\displaystyle {\frac {(x^{2}\cos x+3)(x^{2}(\sin x)'+(x^{2})'\sin x)-(x^{2}\sin x+1)(x^{2}(\cos x)'+(x^{2})'\cos x)}{(x^{2}\cos x+3)^{2}}}\\&&\\&=&\displaystyle {{\frac {(x^{2}\cos x+3)(x^{2}\cos x+2x\sin x)-(x^{2}\sin x+1)(-x^{2}\sin x+2x\cos x)}{(x^{2}\cos x+3)^{2}}}.}\end{array}}$

So, we get

$h'(x)={\frac {(x^{2}\cos x+3)(x^{2}\cos x+2x\sin x)-(x^{2}\sin x+1)(-x^{2}\sin x+2x\cos x)}{(x^{2}\cos x+3)^{2}}}.$

Exercise 4

Evaluate  $\int x{\sqrt {x+1}}~dx.$

First, using the Quotient Rule, we have

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

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

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

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

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

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

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

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

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

So, we have

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