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Rules For Derivative

Learn rules and formulas for derivative, eliminates the use of limit every time and simplifies the calculation of derivative. Practice derivative power rule, product and quotient rule derivative.

Differentiation Formulas

  • 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) \(\dfrac{d}{dt}(2)=0\)

Illustration Questions

If \(y=3^{70}\), then the derivative \(\dfrac{dy}{dx}\) will be

A 5

B \(2x\)

C \(7x^2\)

D 0

×

\(y=3^{70}\) = constant \(\to\) does not depend on \(x\).

\(\Rightarrow\,\dfrac{dy}{dx}=0\)

If \(y=3^{70}\), then the derivative \(\dfrac{dy}{dx}\) will be

A

5

.

B

\(2x\)

C

\(7x^2\)

D

0

Option D is Correct

Derivative Rules

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

Illustration Questions

If \(f(x)=x^7\) then \(f'(x)\) =

A \(18x^9\)

B \(7x^6\)

C \(24x^{25}\)

D \(94x\)

×

\(f(x)=x^7\)

\(\Rightarrow\) \(f'(x)\) \(=7x^6\)

because \(f(x)=x^n\)

\(\Rightarrow\) \(f'(x)\) \(=nx^{n-1}\)     (here \(n=7\))

If \(f(x)=x^7\) then \(f'(x)\) =

A

\(18x^9\)

.

B

\(7x^6\)

C

\(24x^{25}\)

D

\(94x\)

Option B is Correct

Derivative of Constant times a Function

(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.

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

\(=2×4×x^{4-1}=8x^3\)

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

\(=4×1×x^{1-1}=4x^0=4\)

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

\(=6×100\;x^{100-1}\)

\(=600x^{99}\)

Illustration Questions

If \(f(x)=-7x^8\) ,then \(\dfrac{d}{dx}f(x)=\)

A \(74x^8\)

B \(-56x^7\)

C \(28x^2\)

D \(-85x^5\)

×

\(\dfrac{d}{dx}\left(cf(x)\right)=c\,\dfrac{d}{dx}\,(f(x))\)

\(\Rightarrow\,\dfrac{d}{dx}\left(-7x^8\right)=-7\dfrac{d}{dx}x^8\)

\(=-7×8x^7\)

\(=-56x^7\)

If \(f(x)=-7x^8\) ,then \(\dfrac{d}{dx}f(x)=\)

A

\(74x^8\)

.

B

\(-56x^7\)

C

\(28x^2\)

D

\(-85x^5\)

Option B is Correct

Derivative of a Sum of Functions

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

(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.

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

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

\(=2x^1+2×1x^0+0\)

\(=2x+2\)

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

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

\(=3×4x^3+14×3x^2+3×2x^1+0\)

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

Illustration Questions

If  \(h(x)=2x^4+7x^8\) , then  \(h'(x)\) =

A \(42x^2+5x^7\)

B \(8x^3+56x^7\)

C \(15x^2-x\)

D \(47x^5-x^8\)

×

If \(h(x)=f(x)+g(x)\) then \(h'(x)\)\(f'(x)\) + \(g'(x)\)

\(\Rightarrow\,h(x)=(2x^4+7x^8)\) then 

\(h'(x)\) = \(2×4x^3+7×8x^7\)

\(=8x^3+56x^7\)

If  \(h(x)=2x^4+7x^8\) , then  \(h'(x)\) =

A

\(42x^2+5x^7\)

.

B

\(8x^3+56x^7\)

C

\(15x^2-x\)

D

\(47x^5-x^8\)

Option B is Correct

Derivative of Difference of Functions

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

(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.

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

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

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

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

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

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

\(=12×3x^2-4×2x+1-5×5x^4\)

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

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

Illustration Questions

If \(h(x)=25x^4-9x^3+x\), then \(h'(x)\) =

A \(100x^3-27x^2+1\)

B \(25x^7-8x^3+12\)

C \(54x^2-18x\)

D \(27x^{10}+8x^2+7\)

×

\(h(x)=f(x)-g(x)+\phi(x)\)

\(\Rightarrow\) \(h'(x)\)\(f'(x)\) – \(g'(x)\) + \(\phi'(x)\)

\(\Rightarrow\,h(x)=25x^4-9x^3+x\) then 

\(h'(x)\) = \(25×4x^3-9×3x^2+1\)

\(=100x^3-27x^2+1\)

If \(h(x)=25x^4-9x^3+x\), then \(h'(x)\) =

A

\(100x^3-27x^2+1\)

.

B

\(25x^7-8x^3+12\)

C

\(54x^2-18x\)

D

\(27x^{10}+8x^2+7\)

Option A is Correct

Derivative of Product of Functions

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:

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

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

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

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

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

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

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

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

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

Illustration Questions

If \(f(x)=(2x^2+x+3)\) and \(g(x)=(5x+7)\) then \(\dfrac{d}{dx}\left(f(x)\,g(x)\right)=\)

A \(14x^2-28x+55\)

B \(19x^3+45x^2+6\)

C \(2x^3+8x^2+18\)

D \(30x^2+38x+22\)

×

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

\(\Rightarrow\,\dfrac{d}{dx}(2x^2+x+3)\,(5x+7)=(2x^2+x+3)\dfrac{d}{dx}(5x+7)+(5x+7)\dfrac{d}{dx}(2x^2+x+3)\)

\(=(2x^2+x+3)×(5×1+0)+(5x+7)×(2×2x+1+0)\)

\(=(2x^2+x+3)×5+(5x+7)(4x+1)\)

\(=10x^2+5x+15+20x^2+5x+28x+7\)

\(=30x^2+38x+22\)

If \(f(x)=(2x^2+x+3)\) and \(g(x)=(5x+7)\) then \(\dfrac{d}{dx}\left(f(x)\,g(x)\right)=\)

A

\(14x^2-28x+55\)

.

B

\(19x^3+45x^2+6\)

C

\(2x^3+8x^2+18\)

D

\(30x^2+38x+22\)

Option D is Correct

Derivative of Quotient of Functions

 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.

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

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

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

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

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

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

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

Illustration Questions

If \(y=\dfrac{x^3-2x}{x^2+x+1}\) then \(\dfrac{dy}{dx}=\)

A \(\dfrac{x^4+2x^3+5x^2-2}{(x^2+x+1)^2}\)

B \(\dfrac{x^4+8x^3+x+3}{(x^2+5x+6)^2}\)

C \(\dfrac{2x^2-8x+7}{(5x+1)^2}\)

D \(\dfrac{8x^2+7x+3}{(x^2+x+1)^2}\)

×

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

Here,

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

\(g(x)=x^2+x+1\)

\(\Rightarrow\,\dfrac{d}{dx}\left(\dfrac{x^3-2x}{x^2+x+1}\right)=\dfrac{(x^2+x+1)\dfrac{d}{dx}(x^3-2x)-(x^3-2x)\dfrac{d}{dx}(x^2+x+1)}{(x^2+x+1)^2}\)

\(=\dfrac{(x^2+x+1)×(3x^2-2)-(x^3-2x)\,(2x+1)}{(x^2+x+1)^2}\)

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

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

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

If \(y=\dfrac{x^3-2x}{x^2+x+1}\) then \(\dfrac{dy}{dx}=\)

A

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

.

B

\(\dfrac{x^4+8x^3+x+3}{(x^2+5x+6)^2}\)

C

\(\dfrac{2x^2-8x+7}{(5x+1)^2}\)

D

\(\dfrac{8x^2+7x+3}{(x^2+x+1)^2}\)

Option A is Correct

General Power Rule

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

Example:

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

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

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

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

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

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

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

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

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

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

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

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

Illustration Questions

If \(y=\dfrac{2\sqrt x}{\sqrt x+1}\) then \(\dfrac{dy}{dx}=\)

A \(\dfrac{2}{x^2(\sqrt x+1)}\)

B \(\dfrac{1}{\sqrt x(\sqrt x+1)^2}\)

C \(\dfrac{5x}{( x+1)^2}\)

D \(\dfrac{3x^3}{x^2+1}\)

×

\(y=\dfrac{2\sqrt x}{\sqrt x+1}\)

\(\Rightarrow\,\dfrac{dy}{dx}=\dfrac{(\sqrt x+1)\dfrac{d}{dx}(2\sqrt x)-2\sqrt x\dfrac{d}{dx}(\sqrt x+1)}{(\sqrt x+1)^2}\)

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

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

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

If \(y=\dfrac{2\sqrt x}{\sqrt x+1}\) then \(\dfrac{dy}{dx}=\)

A

\(\dfrac{2}{x^2(\sqrt x+1)}\)

.

B

\(\dfrac{1}{\sqrt x(\sqrt x+1)^2}\)

C

\(\dfrac{5x}{( x+1)^2}\)

D

\(\dfrac{3x^3}{x^2+1}\)

Option B is Correct

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