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Quotient Rule

  • Last Updated : 20 May, 2021

In differentiation, Two functions connected by some Operators can also be easily differentiated depending upon the type of operator used. Suppose, A positive sign is used between two functions, the functions can be separately differentiated with the +ve sign there, the same case happens with a negative sign as well. However, the same is not possible for multiplication or a division sign. In multiplication, each function is separately differentiated while taking the other function as a constant, and if the functions are dividing, the rule becomes a bit complicated. Let’s learn about this rule famously known as the quotient rule in differentiation.

What is Quotient Rule?

Let’s suppose two functions are given f(x) and g(x) and both functions are differentiable, that is, the derivative of both functions exists, Then the quotient rule is used to solve the quotient of the functions given.

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Quotient rule states that the derivative of the entire function is the derivative of the numerator times denominator minus the derivative of the denominator times numerator whole divided by the square of the denominator



H'(x)=\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{[f'(x)g(x)-f(x)g'(x)]}{[g(x)]^2}

It is easy to remember the quotient rule as the numerator is the same as that of the product rule just make sure to put a negative sign instead of a positive, also the denominator is the square of the denominator function only.

Proof of Quotient Rule:

There are several ways to prove the quotient rule, one of the ways to prove the rule is by using standard derivative definition and limit properties. The other way is by simply applying product rule and arranging the functions.

Let’s prove the quotient rule by latter method,

To ProveH'(x)=\frac{d}{dx}\frac{f(x)}{g(x)}=\frac{[f'(x)g(x)-f(x)g'(x)]}{[g(x)]^2}

Given, H(x)=f(x)/g(x)

Proof,



f(x)= H(x)g(x)

Applying Product rule here,

f'(x)=H'(x)g(x)+g'(x)H(x)

Put the value of H(x),

H'(x)= \frac{f'(x)-\frac{g'(x)f(x)}{g(x)}}{g(x)}

H'(x)= \frac{f'(x)g(x)-{g'(x)f(x)}}{g(x)^2}

Hence, Proved.

Sample problems

Question 1: Differentiate y= \frac{x^3-x+2}{x^2+5} 

Solution:

Both numerator and denominator functions are differentiable.



Applying quotient rule,

y'=\frac {d}{dx}[\frac{x^3-5+2}{x^2+5}]

y'= \frac{[d/dx(x^3-x+2)(x^2+5)-(x^3-x+2)d/dx(x^2+5)]}{[x^2+5]^2}

y'= \frac{[(3x^2-1)(x^2+5)-(x^3-x+2)(2x)]}{[x^2+5]^2}\\=\frac{(3x^4+15x^2-x^2-5)-(2x^4-2x^2+4x)}{[x^2+5]^2}

y'= \frac{x^4+16x^2-4x-5}{[x^2+5]^2}

Question 2: Differentiate, f(x) = tanx

Solution:

tanx can be written as sinx/cosx

Both the numerator and denominator functions have a possible derivative.

Applying quotient rule,



f'(x)=\frac{(d/dx(sinx))(cosx)-(d/dx(cosx))(sinx)}{cos^2x}

f'(x)= \frac{cosx.cosx-(-sinx)(sinx)}{cos^x}

f'(x)=\frac{cos^2x+sin^2x}{cos^2x}

f'(x)=\frac{1}{cos^2x}

Question 3: Differentiate, f(y)=\frac{8}{y^3}

Solution:

Derivatives of both numerator and denominator are present,

Applying quotient rule,

f'(y)=\frac{(d/dy(8))(x^3)-(d/dy(x^3))(8)}{x^6}

f'(y)=\frac{0.x^3-3x^2(8)}{x^6}



f'(y)=\frac{-24x^2}{x^6}

Question 4: Differentiate, y=\frac{\sqrt{x}+x}{x+2}

Solution:

Both numerator and denominator have possible derivatives.

Applying quotient rule,

y'=d/dx[\frac{\sqrt{x}+x}{x+2}]

y'=\frac{d/dx[\sqrt{x}+x](x+2)-d/dx(x+2)(\sqrt{x}+x)}{[x+2]^2}

y'=\frac{(\frac{1}{2\sqrt{x}}+1)(x+2)-(\sqrt{x}+x)}{[x+2]^2}

y'=\frac{\frac{\sqrt{x}}{2}+\frac{1}{\sqrt{x}}+2-\sqrt{x}}{[x+2]^2}

y'=\frac{\frac{-\sqrt{x}}{2}+\frac{1}{\sqrt{x}}+2}{[x+2]^2}



Question 5: Differentiate, f(x)= ex/x2

Solution:

Both numerator and denominator have possible derivatives,

Applying quotient rule,

f'(x)=[\frac{d/dx(e^x)(x^2)-d/dx(x^2)(e^x)}{x^4}]

f'(x)=\frac{e^x.x^2-2xe^x}{x^4}

Question 6: Differentiate, y=\frac{cosx}{x^2}

Solution:

Both numerator and denominator have possible derivatives,

Applying quotient rule,

y'=\frac{d/dx(cosx)(x^2)-d/dx(x^2)(cosx)}{x^4}

y'=\frac{-sinx(x^2)-(2x)(cosx)}{x^4}

y'=\frac{-(x^2)sinx-(2xcosx)}{x^4}

Question 7: Differentiate, f(p)= p+5/p+7

Solution:

Both numerator and denominator have possible derivatives.

Applying Quotient rule,

f'(p)=d/dx[\frac{p+5}{p+7}]

f'(p)=[\frac{d/dx(p+5)(p+7)-d/dx(p+7)(p+5)}{(p+7)^2}]

f'(p)=[\frac{p+7-p-5}{(p+7)^2}]



f'(p)=[\frac{2}{(p+7)^2}]

Question 8: Differentiate, f(x)=\frac{x^4+3}{\sqrt{x+5}}

Solution:

Both numerator and denominator have possible derivatives

Applying quotient rule,

f'(x)= d/dx[\frac{x^4+3}{\sqrt{x+5}}]

f'(x)=\frac{d/dx(x^4+3)(\sqrt{x+5})-d/dx(\sqrt{x+5})(x^4+3)}{{(x+5)}}

f'(x)=\frac{(4x^3)(\sqrt{x+5})-(\frac{1}{2\sqrt{x+5}})(x^4+3)}{{x+5}}

f'(x)=\frac{(4x^3)({x+5})-(\frac{1}{2})(x^4+3)}{{(x+5)^{3/2}}}

f'(x)=\frac{(8x^3)({x+5})-(x^4+3)}{{2(x+5)^{3/2}}}

f'(x)=\frac{8x^4+40x^3-x^4-3}{{2(x+5)^{3/2}}}

f'(x)=\frac{7x^4+40x^3-3}{{2(x+5)^{3/2}}}




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