Let’s talk about differentiation, a key concept in calculus that you'll definitely encounter on exams like the SAT, ACT, and AP exams. Differentiation helps us find the slope of a function at any given point, essentially showing how fast the function is changing. To do this, we use different rules. First, there's the Power Rule, which makes finding derivatives of terms like $x^n$ easy. Then, the Sum and Difference Rule helps when adding or subtracting functions. For multiplying functions, use the Product Rule, and for division, the Quotient Rule comes into play. Finally, the Chain Rule is essential for composite functions. Mastering these rules is crucial for tackling calculus questions in exams like the SAT, MCAT, and even GRE. In this article, we’ll go through each rule and work through some examples to help you ace those derivative problems!

Elementary Rules of Differentiation

Let’s see the elementary rules of differentiation. The derivative of a function is a concept in mathematics of a real variable that measures the sensitivity to change of the function value (output value) with respect to a change in its argument (input value). They are a part of differential calculus. There are various methods of differentiation. The derivative of a function of a single variable at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. The derivatives are used to find solutions to differential equations

The derivative of a function of a real variable measures the sensitivity to change of the function value with respect to a change in its argument. Derivatives are a fundamental tool of calculus. All functions are functions of real numbers (R) that return real values. 

Differentiation is Linear

In calculus, the derivative of any linear combination of functions equals the same linear combination of the derivatives of the functions; this property is known as linearity of differentiation, the rule of linearity, or the superposition rule for differentiation.

\(\begin{matrix}
f(x+y) = m(x+y) = mx + my = f(x) + f(y)\\
(cf(x))′ = {d\over{dx}}cf(x) = c{d\over{dx}}f(x) = cf′(x)\\
(f(x) + g(x))’ = f’(x) + g’(x)\\
=\lim _{Δ{x}{\rightarrow}0}{f(x+Δx)+g(x+Δx)−(f(x)+g(x))\over{Δx}}\\
=\lim _{Δ{x}{\rightarrow}0}{f(x+Δx)+g(x+Δx)−f(x)−g(x)\over{Δx}}\\
=\lim _{Δ{x}{\rightarrow}0}{f(x+Δx)−f(x)+g(x+Δx)−g(x)\over{Δx}}\\
=\lim _{Δ{x}{\rightarrow}0}({f(x+Δx)−f(x)\over{Δx}} + {g(x+Δx)−g(x)\over{Δx}})\\
=\lim _{Δ{x}{\rightarrow}0}({f(x+Δx)−f(x)\over{Δx}}) + \lim _{Δ{x}{\rightarrow}0}({g(x+Δx)−g(x)\over{Δx}})\\
= f’(x) + g’(x)
\end{matrix}\)

There are some elementary rules of differential calculus. They are as follows:

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 

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

Chain Rule

Chain Rule helps us differentiate composite functions with the number of functions that make up the composition determining how many differentiation steps are necessary. The chain rule states that
\(\begin{matrix}
{d\over{dx}}f(g(x)) = f'(g(x))⋅g'(x).\\
{d\over{dx}}f[g(h(x))] = f'(g(h(x)))⋅g'(h(x))h'(x)
\end{matrix}\)

g and h, make up the composite function f, you have to consider the derivatives g′ and h′ in differentiating f( x).

Example: Find f′( x) if

\(\begin{matrix}
f( x) = (3x^2 + 5x − 2)^8.\\
f'(x) = 8(3x^2 + 5x − 2)^7(6x + 5)
\end{matrix}\)

᠎Inverse Function Rule

Functions f and g are inverses if f(g(x))=x=g(f(x)).
Given an invertible function f(x), the derivative of its inverse function \(f^{−1}(x)\) evaluated at x=a is:
\(\begin{equation*}
\left[f^{-1}\right]'(a)=\frac{1}{f’\left[ f^{-1}(a) \right]}
\end{equation*}\)

᠎Sum and Difference 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.
\(\begin{matrix}
{d\over{dx}} [f(x) + g(x)] = {d\over{dx}}f(x) + {d\over{dx}}g(x)\\
{d\over{dx}} [f(x) – g(x)] = {d\over{dx}}f(x) – {d\over{dx}}g(x)
\end{matrix}\)

Power laws, Polynomials, Quotients, and Reciprocals

Now let’s see how we will differentiate Power functions, polynomials, quotients, and reciprocals

Elementary Power Rule or Polynomial Rule

The Power rule tells us how to differentiate expressions of the form x^n (in other words, expressions with x raised to any power)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.

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.

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

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:

\({d\over{dx}}({f(x)\over{g(x)}}) = {g(x){d\over{dx}}f(x) – f(x){d\over{dx}}g(x)\over{g(x)}^2}\)

Reciprocal Rule

The reciprocal rule is very similar to the quotient rule, except that it can only be used with quotients in which the numerator is a constant. Here is the formula:

Given a function

The reciprocal rule is very similar to the quotient rule, except that it can only be used with quotients in which the numerator is a constant. Here is the formula:
Given a function
\(\begin{matrix}
h(x)=\frac{a}{f(x)}\\
\text{ then its derivative is }\\
h'(x) = -a{f'(x)\over{[f(x)]^2}}
\end{matrix}\)

Generalized Power Rule

The general power rule is a special case of the chain rule. It is useful when finding the derivative of a function that is raised to the nth power. The general power rule states that this derivative is n times the function raised to the (n-1)th power times the derivative of the function.
\({d\over{dx}}f(x)=n.f(x)^{n-1}\times{f'(x)}\)

Derivatives of Exponential and Logarithmic Functions

Now let’s see the differentiating rules for various cases of Logarithmic and Exponential Functions

Derivatives of Exponential Functions of x 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.

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.

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

Proof

According to the definition of the derivative, the differentiation of \(f(x)=x^n\) with respect to x can be written in limited operation form.

\(\begin{equation}\\
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=x^n\\
f(x+h)=(x+h)^n=x^n((1+{h\over{x}})^n)\\
f(x+h)–f(x)=x^n[((1+{h\over{x}})^n)-1]\\
{f(x+h)–f(x)\over{h}}={x^n[((1+{h\over{x}})^n)-1]\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}=\lim _{h{\rightarrow}0}{x^n[((1+{h\over{x}})^n)-1]\over{h}}\\
\lim _{h{\rightarrow}0}[{x^n\over{h}}[(1+{h\over{x}})^n-1]]\\
\text{We use Binomial Expansion to evaluate }(1+{h\over{x}})^n-1\\
\lim _{h{\rightarrow}0}[{x^n\over{h}}\times[1+{n\over{1!}}({h\over{x}})+{n(n-1)\over{2!}}({h\over{x}})^2+{n(n-1)(n-2)\over{3!}}({h\over{x}})^3+ … +-1]]\\
\lim _{h{\rightarrow}0}[{x^n\over{h}}\times[{n\over{1!}}({h\over{x}})+{n(n-1)\over{2!}}({h\over{x}})^2+{n(n-1)(n-2)\over{3!}}({h\over{x}})^3+ …]]\\
\lim _{h{\rightarrow}0}[{x^n\over{h}}\times({h\over{x}})[{n\over{1!}}+{n(n-1)\over{2!}}({h\over{x}})+{n(n-1)(n-2)\over{3!}}({h\over{x}})2+ …]]\\
\lim _{h{\rightarrow}0}[{x^n\over{x}}\times[{n\over{1!}}+{n(n-1)\over{2!}}({h\over{x}})+{n(n-1)(n-2)\over{3!}}({h\over{x}})2+ …]]\\
\lim _{h{\rightarrow}0}[{x^{n-1}}\times[{n\over{1!}}+{n(n-1)\over{2!}}({h\over{x}})+{n(n-1)(n-2)\over{3!}}({h\over{x}})2+ …]]\\
\text{Putting h=0}\\
=[{x^{n-1}}\times[{n\over{1!}}+0]]\\
{d\over{dx}}x^n=n.x^{n-1}
\end{equation}\)

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 raised to the power of x with respect to x is written in the following form in calculus.
\({d\over{dx}}a^x=a^x.\ln{a}\)
Proof
According to the definition of the derivative, the differentiation of f(x)=a^x with respect to x can be written in limited operation form.

\(\begin{equation}\\
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=a^x\\
f(x+h)=a^{x+h}\\
f(x+h)–f(x)=x+h-x=a^{x+h}-a^x=a^x(a^h-1)\\
{f(x+h)–f(x)\over{h}}={a^x(a^h-1)\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}=\lim _{h{\rightarrow}0}{a^x(a^h-1)\over{h}}\\
=a^x\lim _{h{\rightarrow}0}{(a^h-1)\over{h}}
\text{According to the limit rule of exponential function,}\\
\text{ the limit of a raised to the power of h minus 1 by h as}\\
\text{ the value h is closer to zero is equal to natural logarithm of a.}\\
f’(x)={dy\over{dx}}=a^x.\ln{a}
\end{equation}\)

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, 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.

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

Proof

According to the definition of the derivative, the differentiation of f(x)=x with respect to x can be written in limited operation form.

\(\begin{equation}\\
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=e^x\\
f(x+h)=e^{x+h}\\
f(x+h)–f(x)=x+h-x=e^{x+h}-e^x=e^x(e^h-1)\\
{f(x+h)–f(x)\over{h}}={e^x(e^h-1)\over{h}}\\
\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}=\lim _{h{\rightarrow}0}{e^x(e^h-1)\over{h}}\\
=e^x\lim _{h{\rightarrow}0}{(e^h-1)\over{h}}\\
\text{According to the limit rule of exponential function,}\\
\text{ the limit of a raised to the power of h minus 1 by h as }\\
\text{ the value h is closer to zero is equal to natural logarithm of a.}\\
f’(x)={dy\over{dx}}=e^x.\ln{e}\\
\text{But ln(e) = 1}\\
f’(x)={dy\over{dx}}=e^x\
\end{equation}\)

Derivative of Log function

The derivative of a logarithmic function of a variable with respect to itself is equal to its reciprocal.
\({d\over{dx}}{logx}={1\over{x}}\)
Proof
According to the definition of the derivative, the differentiation of f(x)=x with respect to x can be written in limited operation form.

\(\begin{equation}\\
f’(x)={dy\over{dx}}=\lim _{h{\rightarrow}0}{f(x+h)–f(x)\over{h}}
f(x)=logx\\
f(x+h)=log(x+h)\\
f(x+h)–f(x)=log(x+h)-logx=log({x+h\over{x}})=log(1+{h\over{x}})\\
{f(x+h)–f(x)\over{h}}={log(1+{h\over{x}})\over{h}}\\
\text{Let’s put (h\x) as (1\n)}\\
{f(x+h)–f(x)\over{h}}=({1\over{h}}){log(1+{1\over{n}})}=({n\over{x}}){log(1+{1\over{n}})}=({1\over{x}})({log(1+{1\over{n}})})^n\\
\text{As h tends to zero (1\n) tends to}\infty\\
\lim _{h{\rightarrow}\infty}{f(x+h)–f(x)\over{h}}=\lim _{h{\rightarrow}\infty}{(1\over{x}})({log(1+{1\over{n}})})^n={(1\over{x}})\lim _{h{\rightarrow}\infty}({log(1+{1\over{n}})})^n\\
\text{The value of the limit is equal to}\\
\lim\limits_{n \to \infty } {\left( {1 + \frac{1}{n}} \right)^n} = e \approx 2.718281828459 \ldots\\
\left( {{{\log }_a}x} \right)^\prime = \frac{1}{x}{\log _a}e\\
{\log _a}e = \frac{{\ln e}}{{\ln a}} = \frac{1}{{\ln a}}\\
f’(x)={d\over{dx}}{logx}={1\over{x}}
\end{equation}\)

Derivatives of Trigonometric Function

The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. All derivatives of circular trigonometric functions can be found from those of sin(x) and cos(x) by means of the quotient rule applied to functions such as tan(x) = sin(x)/cos(x).

Being able to calculate the derivatives of the sine and cosine functions will enable us to solve many problems such as finding the velocity and acceleration of simple harmonic motion.

Derivatives of Trigonometric Function

The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. All derivatives of circular trigonometric functions can be found from those of sin(x) and cos(x) by means of the quotient rule applied to functions such as tan(x) = sin(x)/cos(x).

Being able to calculate the derivatives of the sine and cosine functions will enable us to solve many problems such as finding the velocity and acceleration of simple harmonic motion.

Derivatives of Hyperbolic Functions

The hyperbolic functions are defined as combinations of the exponential functions e^{x} and \(e^{−x}\).

The basic hyperbolic functions are the hyperbolic sine function and the hyperbolic cosine function. They are defined as follows:

\(\sinh x = \frac{{{e^x} – {e^{ – x}}}}{2},\;\;\cosh x = \frac{{{e^x} + {e^{ – x}}}}{2}.\)

The other hyperbolic functions tanh⁡x, coth⁡x, sechx, cschx are obtained from sinh⁡x and cosh⁡x in exactly the same way as the trigonometric functions tan⁡x, cot⁡x, sec⁡x and cosec⁡x are defined in terms of sin⁡x and cosx.

\(\sinh x = \frac{{{e^x} – {e^{ – x}}}}{2},\;\;\cosh x = \frac{{{e^x} + {e^{ – x}}}}{2}.
\text{sech}\,x = \frac{1}{{\cosh x}};\;\;\text{csch}\,x = \frac{1}{{\sinh x}}\;\left( {x \ne 0} \right).\)

The derivatives of hyperbolic functions can be easily found as these functions are defined in terms of exponential functions. So, the derivatives of the hyperbolic sinx and hyperbolic cosx functions are given by

\(\begin{matrix}
\left( {\sinh x} \right)^\prime = \left( {\frac{{{e^x} – {e^{ – x}}}}{2}} \right)^\prime = \frac{{{e^x} + {e^{ – x}}}}{2} = \cosh x,\;\;\;\left( {\cosh x} \right)^\prime = \left( {\frac{{{e^x} + {e^{ – x}}}}{2}} \right)^\prime = \frac{{{e^x} – {e^{ – x}}}}{2} = \sinh x\\
\left( {\tanh x} \right)^\prime = \left( {\frac{{\sinh x}}{{\cosh x}}} \right)^\prime = \frac{{{{\left( {\sinh x} \right)}^\prime }\cosh x – \sinh x{{\left( {\cosh x} \right)}^\prime }}}{{{{\cosh }^2}x}} = \frac{{\cosh x \cdot \cosh x – \sinh x \cdot \sinh x}}{{{{\cosh }^2}x}} = \frac{{{{\cosh }^2}x – {{\sinh }^2}x}}{{{{\cosh }^2}x}}.\\
\left( {\coth x} \right)^\prime = \left( {\frac{{\cosh x}}{{\sinh x}}} \right)^\prime = \frac{{{{\left( {\cosh x} \right)}^\prime }\sinh x – \cosh x{{\left( {\sinh x} \right)}^\prime }}}{{{{\sinh }^2}x}} = – \frac{{{{\cosh }^2}x – {{\sinh }^2}x}}{{{{\sinh }^2}x}} = – \frac{1}{{{{\sinh }^2}x}} = – {\text{csch}^2}x,\\
\left( {\text{sech}\,x} \right)^\prime = \left( {\frac{1}{{\cosh x}}} \right)^\prime = – \frac{1}{{{{\cosh }^2}x}} \cdot {\left( {\cosh x} \right)^\prime } = – \frac{1}{{{{\cosh }^2}x}} \cdot \sinh x = – \frac{1}{{\cosh x}} \cdot \frac{{\sinh x}}{{\cosh x}} = – \text{sech}\,x\tanh x,\\
\left( {\text{csch}\,x} \right)^\prime = \left( {\frac{1}{{\sinh x}}} \right)^\prime = – \frac{1}{{{\sinh^2}x}} \cdot {\left( {\sinh x} \right)^\prime } = – \frac{1}{{{\sinh^2}x}} \cdot \cosh x = – \frac{1}{{\sinh x}} \cdot \frac{{\cosh x}}{{\sinh x}} = – \text{csch}\,x\coth x\;\;\left( {x \ne 0} \right).
\end{matrix}\)

Derivatives of Special Functions

Special Functions include two functions namely Gamma Function and Riemann Zeta Function. Let’s see how we calculate their derivatives.

Gamma Function

The gamma function Gamma(n) is defined to be an extension of the factorial to complex and real number arguments. It is related to the factorial by

\( \Gamma(n)=(n-1)!\)

The gamma function can be defined as a definite integral for R[z]>0 (Euler’s integral form)

\(\begin{matrix}
\Gamma(z) = \int_0^\infty{t^{(z-1)}}e^{(-t)}dt \\
= 2\int_0^\infty{e^{(-t^2)}t^{(2z-1)}}dt\\
\Gamma(z) = \int_0^1[ln(1/t)]^{(z-1)}dt\\
\end{matrix}\)

Riemann Zeta Function

The Riemann zeta function is an extremely important special function of mathematics and physics that arises indefinite integration and is intimately related with very deep results surrounding the prime number theorem. While many of the properties of this function have been investigated, there remain important fundamental conjectures (most notably the Riemann hypothesis) that remain unproved to this day.

\({\zeta’} z = \frac {d \zeta} {dz} = -\sum_{n \mathop = 2}^\infty \frac { \ln n} {n^z}\)

Derivatives of Integrals

If f is continuous on [a, b], then

\(g(x) = \int_a^xf(t)dt……a\leq{x}\leq{b}\)

is continuous on [a, b], differentiable on (a, b), and

g’(x) = f(x)

Essentially, we’re just taking the derivative of an integral. In other words, the derivative of an integral of a function is just the function. Basically, the two cancel each other out like addition and subtraction. Furthermore, we’re just taking the variable in the top limit of the integral, x, and substituting it into the function being integrated, f(t).

Consider a function of x.

\(\begin{matrix}
{\displaystyle F(x)=\int _{a(x)}^{b(x)}f(x,t)dt}F(x)=\int _{a(x)}^{b(x)}f(x,t)dt\\
\text{ where the functions} {\displaystyle f(x,t)}f(x,t) \text{ and } {\displaystyle {\frac {\partial }{\partial x}}\,f(x,t)}{\displaystyle {\frac {\partial }{\partial x}}\,f(x,t)} \text{ are both continuous in }{\displaystyle a(x)}a(x) \text{ and } {\displaystyle b(x)}b(x) \text{ are both continuous and both have continuous derivatives for } {\displaystyle x_{0}\leq x\leq x_{1}}x_{0}\leq x\leq x_{1}. \text{ Then for } {\displaystyle \,x_{0}\leq x\leq x_{1}}{\displaystyle \,x_{0}\leq x\leq x_{1}}:\\
{\displaystyle F'(x)=f(x,b(x))\,b'(x)-f(x,a(x))\,a'(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}\,f(x,t)\;dt\,.}F'(x)=f(x,b(x))\,b'(x)-f(x,a(x))\,a'(x)+\int _{a(x)}^{b(x)}{\frac {\partial }{\partial x}}f(x,t)dt
\end{matrix}\)

Derivatives to nth Order

The nth derivative refers to any one of a number of higher-level derivatives of a function.

When you take the derivative of function one time, you get the first derivative. Differentiating the new function another time gives you the second derivative. Likewise, a third, fourth or fifth application of the rules of differentiation gives us the third derivative, fourth derivative and fifth derivative, respectively. The nth derivative is a formula for all successive derivatives of a function.

Finding the nth derivative means to take a few derivatives (1st, 2nd, 3rd…) and look for a pattern. If one exists, then you have a formula for the nth derivative. In order to find the nth derivative, find the first few derivatives to identify the pattern. Apply the usual rules of differentiation to a function. Find each successive derivative to arrive at the nth.

Let’s find the nth derivative of \(f(x) = x^n\). Since this function has exponents, use the Power Rule to find the first few derivatives.

The first three derivatives of this function are:

\(\begin{matrix}
f′(x) = nx^{n – 1}\\
f′′(x) = n(n – 1)x^{n – 2}\\
f′′′(x) = n(n – 1) (n – 2)x^{n – 3}
\end{matrix}\)

Each successive derivative adds another layer of (n -), which can be written as:

\(n(n-1) (n-2)…(n – n + 1)\)

We can write the formula for the nth derivative as:

\(f(n) = n(n – 1) (n – 2)…(n – n + 1)x^{n – n}\)

The pattern of successive multiplication is called a factorial and is written as n!

Faà di Bruno’s Formula

If f and g are n-times differentiable, then

\(\begin{matrix}
{\displaystyle {\frac {d^{n}}{dx^{n}}}[f(g(x))]=n!\sum _{\{k_{m}\}}^{}f^{(r)}(g(x))\prod _{m=1}^{n}{\frac {1}{k_{m}!}}\left(g^{(m)}(x)\right)^{k_{m}}}{\frac {d^{n}}{dx^{n}}}[f(g(x))]=n!\sum _{\{k_{m}\}}^{}f^{(r)}(g(x))\prod _{m=1}^{n}{\frac {1}{k_{m}!}}\left(g^{(m)}(x)\right)^{k_{m}}\\
\text{ where } {\displaystyle r=\sum _{m=1}^{n-1}k_{m}}r=\sum _{m=1}^{n-1}k_{m} and \text{ the set } {\displaystyle \{k_{m}\}}\{k_{m}\} \text{ consists of all non-negative integer solutions of the Diophantine equation } {\displaystyle \sum _{m=1}^{n}mk_{m}=n}\sum _{m=1}^{n}mk_{m}=n.
\end{matrix}\)

General Leibniz Rule

If f and g are n-times differentiable, then

\({\displaystyle {\frac {d^{n}}{dx^{n}}}[f(x)g(x)]=\sum _{k=0}^{n}{\binom {n}{k}}{\frac {d^{n-k}}{dx^{n-k}}}f(x){\frac {d^{k}}{dx^{k}}}g(x)}{\frac {d^{n}}{dx^{n}}}[f(x)g(x)]=\sum _{k=0}^{n}{\binom {n}{k}}{\frac {d^{n-k}}{dx^{n-k}}}f(x){\frac {d^{k}}{dx^{k}}}g(x)\)

Derivative Rules Solved Examples

Let’s see some examples of Derivative Rules systems that come in exams.

Solved Example: Find the derivative of \(\frac{\sin x+\cos x}{\sin x-\cos x}\)

Solution:

Let \(f(x)=\frac{\sin x+\cos x}{\sin x-\cos x}\), then by quotient rule,

\(\begin{matrix} f^{\prime}(x)=& \frac{\frac{d}{d x}(\sin x+\cos x) \cdot(\sin x-\cos x)-(\sin x+\cos x) \cdot \frac{d}{d x}(\sin x-\cos x)}{(\sin x-\cos x)^{2}} \\=& \frac{(\cos x-\sin x)(\sin x-\cos x)-(\sin x+\cos x)(\cos x+\sin x)}{(\sin x-\cos x)^{2}} \\=& \frac{-\sin x \cos x-\sin ^{2} x-\cos ^{2} x-\sin x \cos x}{(\sin x-\cos x)^{2}} \\=& \frac{-2 \sin ^{2} x-2 \cos ^{2} x}{(\sin x-\cos x)^{2}}=\frac{-2\left(\sin ^{2} x+\cos ^{2} x\right)}{(\sin x-\cos x)^{2}} \\=& \frac{-2}{(\sin x-\cos x)^{2}}\left[\because \sin ^{2} x+\cos ^{2} x=1\right] \end{matrix}\)

Solved Example: Find the derivative of \(cos^2x\)

Solution:

We solve this by using the Generalized Power Rule. The general power rule is a special case of the chain rule. It is useful when finding the derivative of a function that is raised to the nth power.

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

Here,

\(\begin{matrix}f(x) = cos^2x\\{d\over{dx}}f(x) = {d\over{dx}}cos^2x\\= 2cosx\times{d\over{dx}}(cosx)\\= 2cosx\times(-sinx)\\= -2sinxcosx\\= -sin2x\end{matrix}\)

Solved Example: How do we calculate the derivative of 2x?

Solution:

The derivative of 2x by using the power rule.

\(\begin{matrix}\text{ We have: }\\y = 2x\\\text{ Which is the product of two functions, and so we apply the Product Rule for Differentiation: }\\{d\over{dx}}x^n=n.x^{n-1}\\\text{ Here 2 is constant }\\2{d\over{dx}}x^1=2(1).x^{1-1}\\= 2x^{0}\\= 2\end{matrix}/latex]

Conclusion

In conclusion, understanding the basic differentiation rules is essential for mastering calculus and performing well on exams like the SAT, ACT, and AP. Whether it’s the Power Rule, Product Rule, Quotient Rule, or Chain Rule, each one plays a crucial role in simplifying derivative problems. By practicing these rules and working through examples, you’ll be well-equipped to tackle any differentiation question that comes your way!