Rules of Differentiation in Mathematics

Contents

We present a summary of the rules of differentiation in mathematics, along with a PDF book at the end of the article that includes solved examples and exercises not mentioned in the article, as well as all the rules arranged in a way that is easy for students to understand.

First, you must know that the derivative of a composite term is different from the derivative of a single variable like x or y. A term, for example, would be x+5 or x²+3x.

In the image above, the left side shows the differentiation rules for the variable x (non-chain Rule), and the right side shows the differentiation rules for composite terms (chain Rule). These are some, but not all, of the rules explained. You can refer to them as needed; the following section contains practical exercises.

Derivative of a constant number = zero f(x) = 5 >> f'(x) = 0
Derivative of x is its coefficient. Example: f(x) = 3x >> f'(x) = 3
Derivative of a fraction = [ (Derivative of the numerator × Denominator) – (Derivative of the denominator × Numerator) ] ÷ [Denominator squared]
f(x) = u/z >> f'(x) = (u'.z-z'.u)/z²
Derivative of two terms multiplied by each other = Derivative of the first × Second + Derivative of the second × First
f(xy) = xy >> f'(xy) = x'.y+y'.x)
Derivative of a variable raised to a power
f(x) = xⁿ >> f'(x) = n . x^n-1
Derivative of a term raised to a power
f(u) = uⁿ >> f'(u) = n . u^n-1 . u'
Derivative of ln x
f(x) = ln x >> f'(x) =1/x
Derivative of the exponential function e^x:
f(x) = e^x >> f'(x) =e^x
Derivative of cos (cosine):
f(x) = cos(x) >> f'(x) =-sin(x)
Derivative of sin:
f(x) = sin(x) >> f'(x) =cos(x)
Derivative of a square root = 1 ÷ The square root itself
In fact, it is a special case of the general case for Rule 6 where
√x = x^½ >> (√x)’ = (x^½)’ = x^1-½ = x^-½ = 1/x^½ = 1/√x

Our article is a series of differentiation exercises with solutions. We will provide a variety of exercises so you can solve most differentiation cases:

Differentiation Rules with Examples

Derivative of a Fraction

Find the derivative of the following fraction or term:

$${\frac{d}{dx}\left(\frac{3x+9}{2-x}\right)}$$

Solution: The rule states that the derivative of a fraction = [ (Derivative of the numerator × Denominator) – (Derivative of the denominator × Numerator) ] ÷ [Denominator squared]

$${\frac{d}{dx}(\frac{y}{z})=\frac{y’ . z-z’ . y}{z^2}}$$

We apply the rule to our example

$${=\frac{\frac{d}{dx}\left(3x+9\right)\left(2-x\right)-\frac{d}{dx}\left(2-x\right)\left(3x+9\right)}{\left(2-x\right)^2} :\\
\frac{d}{dx}\left(3x+9\right) = 3\\
\frac{d}{dx}\left(2-x\right) = -1}$$

Now we substitute:

$${=\frac{3\left(2-x\right)-\left(-1\right)\left(3x+9\right)}{\left(2-x\right)^2}\\
=\frac{15}{\left(2-x\right)^2}}$$

Derivative of a Fraction Raised to a Power

For the derivative of a term raised to a power, the rule states: if Y is a term raised to the power n, its derivative is:

$${\frac{d}{dx}(Y^n) = n×Y^{n-1}×Y’}$$

Now we come to the composite example with a fraction to apply what we learned and not forget it. Prove that:

$${\frac{d}{dx}\left(\frac{8x}{5+x}\right)^2\:=\:\frac{640x}{\left(5+x\right)^3}}$$

Solution: Let’s assume 8x/5+x is y. Its derivative would be:

Derivative = 2 × y × y’

$${=2\cdot \frac{8x}{5+x}\frac{d}{dx}\left(\frac{8x}{5+x}\right) : \\
\frac{d}{dx}\left(\frac{8x}{5+x}\right) =\frac{40}{\left(5+x\right)^2}\\
}$$

We learned how to find the derivative of a fraction in the previous example. Now we substitute:

$${=2\cdot \frac{8x}{5+x}\cdot \frac{40}{\left(5+x\right)^2}\\
=\frac{640x}{\left(5+x\right)^3}}$$

Derivative of Two Multiplied Terms

The rule states: The derivative of two multiplied terms = (Derivative of the first × Second) + (Derivative of the second × First)

$${\frac{d}{dx}(y.z)=y’ . z+z’ . y}$$

Here we will mention an example of two terms multiplied by an exponential number e. The number e is treated like any constant, so there are no new rules. However, for the sake of speed and ease, we can remember some rules, such as:

The derivative of e^x = e^x, meaning it remains the same. This is easily proven, but there is no space to mention the proof here.

Example: Find the derivative of the following term

$${\frac{d}{dx}\left(1-xe^x\right)^{\:}\:}$$

Solution:

$${\frac{d}{dx}\left(1-xe^x\right)^{\:}\:\\
=\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(xe^x\right)}$$

The rule states: The derivative of two multiplied terms = (Derivative of the first × Second) + (Derivative of the second × First)

$${\frac{d}{dx}\left(xe^x\right) = 1\cdot e^x+e^xx}$$

Now we substitute into the term

$${=0-\left(e^x+e^xx\right)\\
=-e^xx-e^x}$$

Derivative of sin, cos, and tan

Let’s look at the set of rules before moving on to the exercises:

The derivative of sinx is cosx
The derivative of cosx is -sinx
The derivative of tanx is 1/cos²x
The derivative of cotx is -1/sin²x

Find the derivative of the following term:

$${\frac{d}{dx}\left(1-\sin \left(x\right)\right)}$$

Solution:

$${=\frac{d}{dx}\left(1\right)-\frac{d}{dx}\left(\sin \left(x\right)\right)\\
=0-\cos \left(x\right)\\
=-\cos \left(x\right)}$$

Another example: Find the derivative of the following term:

$${\frac{d}{dx}\left(1-\cos ^2\left(x\right)\right)}$$

As we know from the following trigonometric identity:

$${cos^2x+sin^2x\:=\:1\\
1-cos^2x=sin^2x\:}$$

So the term becomes:

$${\frac{d}{dx}\left(1-\cos \:^2\left(x\right)\right)\:=\:=\frac{d}{dx}\left(\sin \:^2\left(x\right)\right)\\
=2\sin \left(x\right)\frac{d}{dx}\left(\sin \left(x\right)\right)\\
=2\sin \left(x\right)\cos \left(x\right): But\\
\mathrm{Use\:the\:Double\:Angle\:identity}:\quad \:2\sin \left(x\right)\cos \left(x\right)=\sin \left(2x\right)\\
=\sin \left(2x\right)}$$

Find the solution to the following exercise:

$${\frac{d}{dx}\left(\frac{1-cos^2x}{cosx}\right)}$$

Solution:

$${\frac{d}{dx}\left(\frac{1-cos^2x}{cosx}\:\right)=\frac{d}{dx}\left(\frac{sin^2x}{cosx}\:\right)\\
\frac{d}{dx}\left(\sin \:\left(x\right)\tan \:\left(x\right)\right)}$$

We apply the product rule for derivatives

$${\left(f\cdot g\right)’=f\:’\cdot g+f\cdot g’\\
=\frac{d}{dx}\left(\sin \left(x\right)\right)\tan \left(x\right)+\frac{d}{dx}\left(\tan \left(x\right)\right)\sin \left(x\right)\\
\cos \left(x\right)\tan \left(x\right)+\sec ^2\left(x\right)\sin \left(x\right)\\
=\sin \left(x\right)\left(1+\sec ^2\left(x\right)\right)}$$