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
7 years, 4 months ago

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please differentiate e^x

samuel ntshiwa - 7 years, 3 months ago

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Ashutosh Agrahari - 7 years, 3 months ago

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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 - 7 years, 1 month ago

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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 - 6 years, 3 months ago

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nicely explained!

Niliansh Sinha - 6 years, 5 months ago

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