Derivatives of algebraic functions help us find the rate at which a function is changing. These are a set of rules or formulas that make it easier and faster to differentiate (or find the derivative of) algebraic expressions.

An algebraic function is any function that involves operations like addition, subtraction, multiplication, division, and powers or roots (like square roots or cube roots) of variables. These functions come from solving polynomial equations.

By using derivatives of algebraic functions, we can solve problems in calculus and also find solutions to many types of differential equations. In this topic, we also learn how to use formulas, solve problems step by step, and sometimes even look at graphs to understand how the derivative behaves.

᠎Sum Rule

The Sum rule says the derivative of a sum of functions is the sum of their derivatives. The Difference rule says the derivative of a difference of functions is the difference of their derivatives.

\({d\over{dx}} [f(x) + g(x)] = {d\over{dx}}f(x) + {d\over{dx}}g(x)\)

Difference Rule

The Difference rule says the derivative of a difference of functions is the difference of their derivatives.

\({d\over{dx}} [f(x) – g(x)] = {d\over{dx}}f(x) – {d\over{dx}}g(x)\)

Product Rule

Sometimes we are given functions that are actually products of other functions. This means, two functions multiplied together. A special rule, the product rule, exists for differentiating products of two (or more) functions.

If y = uv then 

\({dy\over{dx}} = u{dv\over{dx}} + v{du\over{dx}}\)

Quotient Rule

A special rule, the quotient rule, exists for differentiating quotients of two functions. Functions often come as quotients, by which we mean one function divided by another function. There is a formula we can use to differentiate a quotient – it is called the quotient rule.

If f and g are both differentiable, then:

\({d\over{dx}}({f(x)\over{g(x)}}) = {g(x){d\over{dx}}f(x) – f(x){d\over{dx}}g(x)\over{g(x)}}\)

Derivatives of Other Functions

The exponential function is a mathematical function denoted by \(f(x)=\exp\) or \(e^{x}\). In addition, an exponential function possesses a constant as its base and a variable as its exponent. An exp function in mathematics is expressed as \(f\left(q\right)=f\left(b\right)=x^b\), where “b” stands for the variable and “x” denotes the constant which is also termed the base of the function. The Logarithmic function is the inverse of the exponential function. Logarithmic functions in mathematics are an operator which will help you exactly calculate the exponent that will satisfy the exponential equation. The logarithmic function equation is as shown, \(c=\log_{b}a\) for a>0 such that b>0 and \(b\ne1\).

Derivatives of Exponential Functions

Let’s see how we can calculate the derivative of exponential functions.

Derivatives of Exponential Functions of x by Power Rule

The derivative of an exponential function, which contains a variable as a base and a constant as power, is called the constant power derivative rule.

x and n are literals and they represent a variable and a constant. They form an exponential term x^n. The derivative of x is raised to the power n is written in mathematical form as follows.

\({d\over{dx}}x^n=n.x^{n-1}\)

Derivatives of Exponential Functions of a by Power Rule

The derivative of an exponential term, which contains a variable as a base and a constant as power, is called the constant power derivative rule.

Suppose a and x represent a constant and a variable respectively then the exponential function is written as \(a^x\) in mathematics. The derivative of a raise to the power of x with respect to x is written in the following form in calculus.

\({d\over{dx}}a^x=a^x.\ln{a}\)

Derivatives of Exponential Functions of e by Power Rule

The derivative of an exponential term, which contains a variable as a base and a constant as power, is called the constant power derivative rule.

Assume that x is a variable, then the natural exponential function is written as ex in mathematical form. The derivative of the ex function with respect to x is written in the following mathematical form.

\({d\over{dx}}e^x=e^x\)

Derivative of Logarithmic Function

The derivative of a logarithmic function of the variable with respect to itself is equal to its reciprocal.

\({d\over{dx}}{logx}={1\over{x}}\).

Derivatives of logarithmic functions are used to find out solutions to differential equations.

Derivative of Square Root

The derivative of the square root of x can be found using the power rule. Here the power of x is ½. The derivative of x is raised to the power n is written in mathematical form as follows.

\({d\over{dx}}x^n=n.x^{n-1}\)

\({d\over{dx}}x^{(½)}=(½).x^{(½)-1}=(½).x^{(-½)}\)

Solved Examples based on Derivatives of Algebraic Functions

Here are some solved examples on Derivatives of Algebraic Functions.

Example 1: \(\dfrac{d}{dx}\left(x^{47}\right)\)

Solution:

\(f(x)=\dfrac{d}{dx}\left(x^{47}\right)\)

Using Derivatives of Exponential Functions of x by Power Rule, we get

\({d\over{dx}}x^n=n.x^{n-1}\)

\(\dfrac{d}{dx}\left(x^{47}\right) = 47x^{46}\)

Example 2: \(f(x) = 2\pi\)

Solution:

Using Derivative of a Constant Rule, we get

\({d\over{dx}}k=0\)

\(\begin{matrix}
\dfrac{d}{dx}(2\pi) &= \dfrac{d}{dx}(\text{constant}) \\
= 0 \
\end{matrix}
\)

Example 3: \(f(x) = \dfrac{2}{3}x^9\)

Solution:

\(f(x) = \dfrac{2}{3}x^9\)

Using Derivatives of Exponential Functions of x by Power Rule, we get

\(\begin{matrix}
\dfrac{d}{dx}\left( \frac{2}{3}x^9\right) &= \frac{2}{3} \dfrac{d}{dx}\left(x^9 \right) \\
= \frac{2}{3}\left(9 x^{9-1} \right) \\
= \frac{2}{3}(9) \left(x^8 \right) \\
= 6x^8
\end{matrix}\)

Example 4: Find the derivative of the function
f(x) = ∛x − 1/√x

Solution:
First, rewrite the function using exponents:
f(x) = x^(1/3) − x^(-1/2)

Now differentiate using the power rule:
d/dx (x^(1/3)) = (1/3) x^(-2/3)
d/dx (-x^(-1/2)) = (1/2) x^(-3/2)

Answer: f '(x) = (1/3)x^(-2/3) + (1/2)x^(-3/2).

Example 5: Find the derivative of f(x) = e^(1 + x)

Solution:
We use the rule:
d/dx (e^u) = e^u · (du/dx)

Here, u = 1 + x, so du/dx = 1

Therefore,
d/dx (e^(1 + x)) = e^(1 + x) · 1 = e^(1 + x)

Answer: f '(x) = e^(1 + x)

Example 6: \( f(x) = \sqrt{x}\left(x^2 – 8 + \dfrac{1}{x} \right)\)

Solution: \( f(x) = \sqrt{x}\left(x^2 – 8 + \dfrac{1}{x} \right)\)

\(\begin{matrix}
\sqrt{x}\left(x^2 – 8 + \frac{1}{x} \right) &= x^{\frac{1}{2}}x^2 – 8x^{\frac{1}{2}} + x^{\frac{1}{2}}x^{-1} \\
&= x^{\frac{5}{2}}\, – 8x^{\frac{1}{2}}\, + x^{-\frac{1}{2}}
\end{matrix}\)

Using Derivatives of Exponential Functions of x by Power Rule, we get

\(\begin{matrix}
\dfrac{d}{dx} \left(\sqrt{x}\left(x^2 – 8 + \frac{1}{x} \right) \right) &= \dfrac{d}{dx}\left(x^{\frac{5}{2}}\right) – 8\dfrac{d}{dx}\left(x^{\frac{1}{2}} \right) + \dfrac{d}{dx} \left( x^{-\frac{1}{2}} \right) \\
= \frac{5}{2} x^{\left(\frac{5}{2}\, – 1 \right)} – 8 \left(\frac{1}{2} x^{\left(\frac{1}{2}\, – 1 \right)} \right) + \left(-\frac{1}{2} \right)x^{\left(-\frac{1}{2}\, -1 \right)} \\
= \frac{5}{2}x^{\frac{3}{2}}\, – 4 x^{-\frac{1}{2}} \,- \frac{1}{2} x^{-\frac{3}{2}} \end{matrix}\)

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