• Finding derivatives directly from the definition for complicated functions is not very easy and will often involve very complex limits.
• There are some formulas which if remembered, eliminates the use of limit everytime and simplifies the calculation of derivative.

The constant function is an expression which does not involve \(x\) or does not change, if we change the value of \(x\).

Example:

(1)  \(\dfrac{d}{dx}(c)=0\)      where, \('c'\) is a constant.

The derivative of constant function is 0.

(2) \(\dfrac{d}{dx}(\pi)=0\)

(3) \(\dfrac{d}{dt}(3)=0\)

(4) \(\dfrac{d}{dt}(2)=0\)

(1)  \(\dfrac{d}{dx}(x)=1\)

(2)  \(\dfrac{d}{dx}(x^n)=nx^{n-1}\) where \(n\) is any positive integer.

\(\dfrac{d}{dx}(x)=\dfrac{d}{dx}\) \(x^1\) = \(1x^0\)

\(=1×1=1\)

Example : \(\dfrac{d}{dx}(x^{20})=20x^{19}\)

(3) \(\dfrac{d}{dx}(x^{15})=15x^{14}\)

(4) \(\dfrac{d}{dx}(x^{100})=100x^{99}\)

(5) \(\dfrac{d}{dx}(x^{250})=250x^{249}\)

  \(\dfrac{d}{dx}\left(f(x)-g(x)\right)=\dfrac{d}{dx}f(x)-\dfrac{d}{dx}\left(g(x)\right)\)

(The derivative of difference of two function is the difference of derivative).

• This assumes that \('f'\) and \('g'\) are differentiable.
• Consider examples to understand this.

\((1)\;\dfrac{d}{dx}(x^5-x^4+x^3-x^2+1)\)

\(=\dfrac{d}{dx}(x^5)-\dfrac{d}{dx}(x^4)+\dfrac{d}{dx}(x^3)-\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(1)\)

\(=5x^4-4x^3+3x^2-2x^1+0\)

\(=5x^4-4x^3+3x^2-2x\)

\((2)\;\dfrac{d}{dx}(12x^3-4x^2+x-5x^5)\)

\(=\dfrac{d}{dx}(12x^3)-\dfrac{d}{dx}(4x^2)+\dfrac{d}{dx}(x)-\dfrac{d}{dx}(5x^5)\)

\(=12×3x^2-4×2x+1-5×5x^4\)

\(=36x^2-8x+1-25x^4\)

\(=-25x^4+36x^2-8x+1\)

Product Rule

\(\dfrac{d}{dx}\left(f(x)\,g(x)\right)=f(x)\,\dfrac{d}{dx}g(x)+g(x)\dfrac{d}{dx}f(x)\)

• The derivative of product of two functions is first function times the derivative of second added to the second function times the derivative of first function.
• This assumes that \('f'\) and \('g'\) are both differentiable.
• Since the formula of product rule is symmetric about \(f\) and \(g\) we can choose any of the two function as \(f \) or \(g\).
• Let's take an example to understand this.

\((1)\;\dfrac{d}{dx}(x+2)(x^2-3x)\)

The given function is a product of two functions, \((x+2)\) and \((x^2-3x)\) .

We can take any function as \(f(x)\) and \(g(x)\)

Let \((x+2)=f(x)\;\text{and}\;(x^2-3x)=g(x)\)

Now, applying product rule:

\(\dfrac{d}{dx}(x+2)(x^2-3x)\)

\(=(x+2)\dfrac{d}{dx}(x^2-3x)+(x^2-3x)\dfrac{d}{dx}(x+2)\)

\(=(x+2)\left[\dfrac{d}{dx}(x^2)-\dfrac{d}{dx}(3x)\right]+(x^2-3x)\left[\dfrac{d}{dx}(x)+\dfrac{d}{dx}(2)\right]\)

\(=(x+2)[2x-3x^0]+(x^2-3x)[x^0+0]\)

\(=(x+2)[2x-3]+(x^2-3x)[1]\)

\(=(x+2)[2x-3]+x^2-3x\)

\(=x(2x-3)+2(2x-3)+x^2-3x\)

\(=2x^2-3x+4x-6+x^2-3x\)

\(=3x^2-2x-6\)

(1)  \(\dfrac{d}{dx}\left(cf(x)\right)=c\dfrac{d}{dx}\left(f(x)\right)\)  where \('c'\) is a constant

(The derivative of constant times a function is constant times the derivative of function).

• Consider examples to understand this.

\((1)\;\dfrac{d}{dx}(2x^{4})=2\dfrac{d}{dx}(x^4)\)    \(\left[\dfrac{d}{dx}x^n=nx^{n-1}\right]\)

\(=2×4×x^{4-1}=8x^3\)

\((2)\;\dfrac{d}{dx}(4x)=4×\dfrac{d}{dx}(x)\)

\(=4×1×x^{1-1}=4x^0=4\)

\((3)\;\dfrac{d}{dx}(6x^{100})=6×\dfrac{d}{dx}(x^{100})\)

\(=6×100\;x^{100-1}\)

\(=600x^{99}\)

  \(\dfrac{d}{dx}\,\left(f(x)+g(x)\right)=\dfrac{d}{dx}\,\left(f(x)\right)+\dfrac{d}{dx}\,\left(g(x)\right)\)

(The derivative of a sum of function is the sum of derivatives)

• This assumes that \(f\) and \(g\) are both differentiable.
• It can be extended to more than two function.

\(\Rightarrow\) \((f+g+h)\)\('\)= \(f'+g'+h'\) 

• Consider examples to understand this.

\((1)\;\dfrac{d}{dx}(x^2+2x+1)\)

\(=\dfrac{d}{dx}(x^2)+\dfrac{d}{dx}(2x)+\dfrac{d}{dx}(1)\)

\(=2x^1+2×1x^0+0\)

\(=2x+2\)

\((2)\;\dfrac{d}{dx}\left(3x^4+14x^3+3x^2+2\right)\)

\(=\dfrac{d}{dx}(3x^4)+\dfrac{d}{dx}(14x^3)+\dfrac{d}{dx}(3x^2)+\dfrac{d}{dx}(2)\)

\(=3×4x^3+14×3x^2+3×2x^1+0\)

\(=12x^3+42x^2+6x\)

 Ratio Rule or the Quotient Rule

\(\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)=\dfrac{g(x)\dfrac{d}{dx}f(x)-f(x)\dfrac{d}{dx}g(x)}{(g(x))^2}\)

• Derivative of quotient is the denominator times the derivative of numerator subtracts the numerator times the derivative of denominator, all of this divided by square of denominator.
• This assumes that \('f'\) and \('g'\) are differentiable and \(g\neq 0\).
• We can remember this formula as-

\(\dfrac{d}{dx}\left(\dfrac{\text{High}}{\text{Low}}\right)\)

\(=\dfrac{\text{Low}\frac{d}{dx}\text{(High)}-\text{(High)}\frac{d}{dx}(\text{Low})}{(\text{Low})^2}\)

"Low D high minus high D low cross the line and square the low"

• Consider an example to understand this.

\(\dfrac{d}{dx}\left(\dfrac{x^2+2}{x+1}\right)\)

\(=\dfrac{(x+1)\frac{d}{dx}(x^2+2)-(x^2+2)\frac{d}{dx}(x+1)}{(x+1)^2}\)

\(=\dfrac{(x+1)\left[\frac{d}{dx}(x^2)+\frac{d}{dx}(2)\right]-(x^2+2)\left[\frac{d}{dx}(x)+\frac{d}{dx}(1)\right]}{(x+1)^2}\)

\(=\dfrac{(x+1)[2x+0]-(x^2+2)[1+0]}{(x+1)^2}\)

\(=\dfrac{(x+1)(2x)-(x^2+2)}{(x+1)^2}\)

\(=\dfrac{2x^2+2x-x^2-2}{(x+1)^2}\)

\(=\dfrac{x^2+2x-2}{(x+1)^2}\)

\(\dfrac{d}{dx}(x^n)=nx^{n-1}\) for all values of \(n\) .

Example:

\((1)\;\dfrac{d}{dx}\left(\dfrac{1}{\sqrt x^3}\right)\)

\(=\dfrac{d}{dx}\left(x^\frac{-3}{2}\right)\)

\(=\dfrac{-3}{2}x^{-3/2-1}\)

\(=\dfrac{-3}{2}x^{-5/2}\)

\((2)\;\dfrac{d}{dx}\left(\dfrac{2\sqrt x}{x^{\frac{4}{5}}}\right)\)

\(=\dfrac{d}{dx}\left(2\dfrac{x^{\frac{1}{2}}}{x^{\frac{4}{5}}}\right)\)

\(=\dfrac{d}{dx}\left(2x^{\frac{1}{2}-\frac{4}{5}}\right)\)

\(=\dfrac{d}{dx}\left(2x^{\frac{5-8}{10}}\right)\)

\(=\dfrac{d}{dx}\left(2x^{\frac{-3}{10}}\right)\)

\(=2×\left(\dfrac{-3}{10}\right)x^{\frac{-3}{10}-1}\)

\(=\dfrac{-3}{5}x^{\frac{-3-10}{10}}\)

\(=\dfrac{-3}{5}x^{\frac{-13}{10}}\)