Rules of Differentiation

Definition and Technique

Differentiation Rules are formulas that allow us to take a complicated derivative and break it up into smaller pieces so we can apply more basic differentiation rules.

The Product Rule

The product rule states that for the functions uu and vv:

ddx(uv)=udvdx+vdudx \frac{d}{dx} \left(u\cdot v\right) = u\cdot\frac{dv}{dx} + v\frac{du}{dx}

For example:

Let u=x2u = x^2 and v=exv = e^x . If f(x)=uv f(x) = u\cdot v , find f(x) f'(x) .

ddx(x2ex)=x2uexdvdx+exv2xdudx=ex(x2+2x) \begin{aligned} \frac{d}{dx}\left(x^2 \cdot e^x \right) &= \underbrace{x^2}_u \cdot \underbrace{e^x}_{\frac{dv}{dx}} + \underbrace{e^x}_v \cdot \underbrace{2x}_\frac{du}{dx} \\ &= e^x\left(x^2+2x\right)\, _\square \end{aligned}

The Quotient Rule

The quotient rule is useful when you have a rational expression. In Lagrange's notation:

(pq)=qppqq2 \left(\frac{p}{q}\right)' = \frac{q \cdot p' - p\cdot q'}{q^2}

Find f(x) f'(x) if f(x)=sinxx3 f(x)=\frac{\sin x}{x^3} .

Let p=sinx p=\sin x and let q=x3 q=x^3 . Thus, p=cosxp'=\cos x and q=3x2q'=3x^2, so:

f(x)=(x3)(cosx)(sinx)(3x2)x6=x2(xcosx3sinx)x6=xcosx3sinxx4 \begin{aligned} f'(x)&=\frac{\left(x^3\right)\left( \cos x \right) - \left(\sin x\right)\left(3x^2\right)}{x^6} \\ &= \frac{x^2\left(x \cos x - 3 \sin x\right)}{x^6} \\ &= \frac{x \cos x - 3 \sin x}{x^4} \, _\square \end{aligned}

The Chain Rule

The Chain Rule is needed for composite functions. Consider f(x)=g(h(x)). f(x)=g(h(x)).

The chain rule explains how we can relate f(x) f'(x) to the derivative of gg and h h . Specifically:

f(x)=g(h(x))h(x). f'(x) = g'(h(x)) \cdot h'(x).

Find the derivative of f(x)=cosx f(x) = \sqrt{\cos x} .

Let g(x)=x g(x) = \sqrt{x} , and h(x)=cosx h(x) = \cos x . Taking the derivatives of these functions, we get:

g(x)=12x g'(x)=\frac{1}{2 \sqrt{x}} h(x)=sinx h'(x) = -\sin x So, according to the chain rule:

f(x)=12cosxg(h(x))sinxh(x)=sinx2cosx \begin{aligned} f'(x) &= \underbrace{\frac{1}{2\sqrt{\cos x}}}_{g'(h(x))} \cdot \underbrace{-\sin x}_{h'(x)} \\ &= -\frac{\sin x}{2\sqrt{\cos x}} \, _\square \end{aligned}

Applications and Extensions

Let's consider a more complicated example:

Find the derivative of f(x)=x2+1x2+3 \displaystyle f(x) = \frac{\sqrt{x^2+1}}{x^2+3} .

First, notice that this is rational function, so we'll need to use the quotient rule. The derivative of the denominator ( x2+3 x^2+3 ) is simply 2x 2x . In order to use the quotient rule, however, we'll also need to know the derivative of the numerator, which we can't find directly.

In order to find the derivative of the numerator, x2+1 \sqrt{x^2+1} , we'll use the chain rule: ddx(x2+1)=12x2+1ddx(x2+1)=12x2+12x=xx2+1 \begin{aligned} \frac{d}{dx}\left( \sqrt{x^2+1} \right) &= \frac{1}{2\sqrt{x^2+1}}\cdot \frac{d}{dx}\left( x^2 + 1 \right) \\ &= \frac{1}{2\sqrt{x^2+1}}\cdot 2x \\ &= \frac{x}{\sqrt{x^2+1}} \end{aligned} Now that we know the derivatives of both the numerator and the denominator, we can proceed to use the quotient rule.

f(x)=(x2+3)(xx2+1)x2+1(2x)(x2+3)2 f'(x) = \frac{\left(x^2+3\right)\left( \dfrac{x}{\sqrt{x^2+1}} \right) - \sqrt{x^2+1}\left(2x\right)}{\left( x^2 +3 \right)^2}

Go through each term, and make sure you understand why it's there. You may want to refer again to the statement of the quotient rule above.

Now we should simplify, so factoring out 1x2+1 \frac{1}{\sqrt{x^2+1}} , we get:

f(x)=1x2+1((x2+3)(x)(x2+1)(2x)(x2+3)2) f'(x) = \frac{1}{\sqrt{x^2+1}} \left(\frac{\left(x^2+3\right)\left( x \right) - \left(x^2+1\right) \left(2x\right)}{\left( x^2 +3 \right)^2}\right)

Multiply and combine like terms:

f(x)=1x2+1(x3+3x2x32x(x2+3)2)=1x2+1(xx3(x2+3)2) \begin{aligned} f'(x) &= \frac{1}{\sqrt{x^2+1}} \left(\frac{x^3+3x-2x^3-2x}{\left( x^2 +3 \right)^2}\right) \\ &= \frac{1}{\sqrt{x^2+1}} \left(\frac{x-x^3}{\left( x^2 +3 \right)^2}\right) \, _\square \end{aligned}

Note by Arron Kau
5 years, 2 months ago

No vote yet
1 vote

</code>...<code></code> ... <code>.">   Easy Math Editor

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link]( link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in </span>...<span></span> ... <span> or </span>...<span></span> ... <span> to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}


Sort by:

Top Newest

please differentiate e^x

samuel ntshiwa - 5 years, 2 months ago

Log in to reply


Ashutosh Agrahari - 5 years, 1 month ago

Log in to reply

differentiation of exe^{x} is exe^{x} only . we can do it by log method too. let y= exe^{x} ..........> Eq1 taking log both sides we get log y = log exe^{x} since log and exponential are inversely propotional logexe^{x}is equal to x then by differentiating both sides we get ,
1y\frac{1}{y} . dydx\frac{dy}{dx} = 1 ( w.k.t dlogdx\frac{d log}{dx} = 1x\frac{1}{x} ) \ (\frac{dy}{dx}) = 1y\frac{1}{y} . 1 =ex e^{x} .1 = exe^{x} ( from eq 1 ) . hence differentiated hope u got your answer

Puja Shree - 5 years ago

Log in to reply

We can use the fact that: 1x=Lnx(+c)\int \frac{1}{x} = Lnx (+c) to solve the following differential equation: dydx=y\frac{dy}{dx} = y 1ydy=1dx Lny=x+c\Rightarrow\int \frac{1}{y} dy = \int 1 dx \Rightarrow\ Ln y = x +c  y=ex+c\therefore\ y = e^{x+c}  y=ex    dydx=ex\therefore\ y= e^x \implies\frac{dy}{dx} = e^x

Curtis Clement - 4 years, 2 months ago

Log in to reply

nicely explained!

Niliansh Sinha - 4 years, 3 months ago

Log in to reply


Problem Loading...

Note Loading...

Set Loading...