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How to Use Quotient Rule?

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  • Last Updated : 24 May, 2022
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The quotient rule is an important rule in the concept of derivatives. To find the derivatives of complex fractions this quotient rule is used. This article is about the quotient rule in derivatives, and how it is applied and used. The quotient rule helps to find the derivative of complex fractions very easily. It is used to find the derivative when the problem is given in fraction form i.e. in the numerator and denominator form.

What is the Quotient Rule?

The quotient rule is the rule for finding the derivative of the functions which is in the fraction form of two differentiable functions.

The quotient rule states that the derivative of the quotient is equal to the ratio of subtraction of denominator function multiplied by derivative of numerator function and numerator function multiplied by derivative of denominator function to the square of the denominator.

The quotient rule is applied when we have to find the derivative of a function of the form first function divided by the second function, which says that the derivative of the quotient is equal to the subtraction of the second function multiplied by the derivative of the first function and first function multiplied by derivative of the second function divided by the square of the second function.

How to Use Quotient Rule?

The Quotient rule formula can be used in the following different ways,

\dfrac{d}{dx}\left[\dfrac{1^{\text{st}}\,\text{function}}{2^{\text{nd}}\,\text{function}}\right] = \dfrac{2^{\text{nd}} \text{function}.\dfrac{d}{dx}[1^{\text{st}}\,\text{function}] - 1^{\text{st}}\,\text{function}.\dfrac{d}{dx}[2^{\text{nd}} \text{function}] }{[2^{\text{nd}}\, \text{function}]^2}


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


\dfrac{d}{dx}\left[\dfrac{u}{v}\right] = \dfrac{v.u' - u.v'}{v^2}           

where, u and v are functions of x.

Sample Questions

Question 1: Find  \dfrac{d}{dx}\left[\dfrac{3x^2+ 6x+2}{x+2}\right]  .


 \dfrac{d}{dx}[\dfrac{3x^2+ 6x+2}{x+2}] = \dfrac{(x+2)\dfrac{d}{dx}(3x^2+6x+2)-(3x^2+6x+2)\dfrac{d}{dx}(x+2)}{(x+2)^2}

= [(x + 2)(6x+6) – (3x2+6x+2)(1)]/(x + 2)2

= [(6x2+6x+12x+12) – (3x2+6x+2)]/(x + 2)2

= (3x2+12x+10)/(x + 2)2

Question 2: Find \dfrac{d}{dx}[\dfrac{sin x}{x^2}]


\dfrac{d}{dx}[\dfrac{sin x}{x^2}] =  \dfrac{x^2.\dfrac{d}{dx}[sin x] - sin x.\dfrac{d}{dx}[x^2] }{[x^2]^2}  

= (x2cosx – 2xsinx)/x

= x.(xcosx – 2sinx)/x4

= (xcosx – 2sinx)/x3 

Question 3: Find \dfrac{d}{dx}[\dfrac{cos(x^2) }{x}]            


\dfrac{d}{dx}[\dfrac{cos (x^2)}{x}] =  \dfrac{x.\dfrac{d}{dx}[cos(x^2 )] - cos(x^2) .\dfrac{d}{dx}[x] }{(x)^2}      

= [-2x2sin(x2) – cos(x2)]/x

Question 4: Find \dfrac{d}{dx}[\dfrac{2x}{5x^2+x+7}]


\dfrac{d}{dx}[\dfrac{2x}{5x^2+x+7}] = \dfrac{(5x^2+x+7).\dfrac{d}{dx}(2x)-(2x).\dfrac{d}{dx}(5x^2+x+7)}{[5x^2+x+7]^2}

= [(5x2+x+7)(2) – (2x)(10x+1)]/(5x2+x+7)2

= [(10x2+2x+14)-(20x2+2x)]/(5x2+x+7)2

= [14-10x2)]/(5x2+x+7)2

Question 5: Find the derivative of  y = (ex + log x)/sin3x


By quotient rule,

\dfrac{d}{dx}[\dfrac{e^x + log x}{sin3x}] = \dfrac{sin3x.\dfrac{d}{dx}[e^x + log x]-(e^x + log x).\dfrac{d}{dx}[sin3x]}{(sin3x)^2}            

= [sin3x . {ex + (1/x)} – (ex + log x)(3cos3x)]/sin23x

= [{ex + (1/x)}.sin3x – 3(ex + log x)cos3x ]/sin23x

Question 6: Find the derivative of  y = (ex + e-x)/ (ex – e-x).


\dfrac{dy}{dx} = \dfrac{d}{dx}[\dfrac{(e^x + e^{-x}) }{(e^x - e^{-x})}]

= \dfrac{d}{dx}[\dfrac{(e^x - e^{-x}).\dfrac{d}{dx}(e^x + e^{-x})-(e^x + e^{-x}).\dfrac{d}{dx}(e^x - e^{-x})  }{(e^x - e^{-x})^2}]  

= [(ex – e-x)(ex – e-x) – (ex + e-x)(ex + e-x)]/(ex – e-x)

= [(ex – e-x)2 – (ex + e-x)2]/(ex – e-x)2

= 4/(ex – e-x) 

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