How to find derivative using quotient rule

Derivatives are mathematical processes that represent a function's instantaneous rate of change at a specific point. The quotient rule provides a formula for finding the derivative of a function that is a quotient of two functions. This rule simplifies the process of finding derivatives of complex fractions and enables efficient differentiation of quotient functions.

Different rules of differentiation
Different rules of differentiation

You can see here for more details of other differentiation rules.

Quotient rule mathematically

The quotient rule states that if you have a function that is a quotient of two functions, such as f(x)g(x)\frac{f(x)}{g(x)} then we can the derivative of such expression by the quotient rule:

ddx(f(x)g(x))=f(x)g(x)g(x)f(x)(g(x))2\frac{d}{{dx}} \left(\frac{{f(x)}}{{g(x)}}\right) = \frac{{f'(x) \cdot g(x) - g'(x) \cdot f(x)}}{ ({g(x))^2}}

Explanation: The derivative of the quotient can be found by taking the derivative of the numerator, f(x)f'(x), multiplied by the denominator, g(x)g(x), minus the derivative of the denominator, g(x)g'(x), multiplied by the numerator, f(x)f(x). Finally, divide this result by the square of the denominator, (g(x))2(g(x))^{2}

Examples of the quotient rule

Let's explore some examples that demonstrate the step-by-step application of the quotient rule and illustrate how it functions in differentiation.

Example 1

Find the derivative of the function. h(x)=3x2+2xx+1.h(x) = \frac{{3x^{2} + 2x}}{{x + 1}}.

Solution: To differentiate this function using the quotient rule, we apply the quotient rule:

h(x)=ddx(f(x)g(x))=f(x)g(x)g(x)f(x)(g(x))2h'(x)=\frac{d}{{dx}} \left(\frac{{f(x)}}{{g(x)}}\right) = \frac{{f'(x) \cdot g(x) - g'(x) \cdot f(x)}}{ ({g(x))^2}}

Let's consider f(x)=3x2+2xf(x) = 3x^{2} +2x and g(x)=x+1g(x) = x+1 , then

h(x)=(ddx(3x2+2x))(x+1)(ddx(x+1))(3x2+2x)(x+1)2h'(x)= \frac{(\frac{d}{{dx}} {(3x^{2}+2x)) \cdot (x+1) - (\frac{d}{{dx}} (x+1)) \cdot (3x^{2}+2x)}}{ ({x+1)^2}}
h(x)=(6x+2)(x+1)(1)(3x2+2x)(x+1)2=3x2+6x+2(x+1)2 h'(x) = \frac{{(6x+2) \cdot (x+1) - (1) \cdot (3x^{2}+2x)}}{ ({x+1)^2}} = \frac{3x^{2}+6x+2}{(x+1)^{2}}

Example 2

Find the derivative of h(x)=tan(x)=sin(x)cos(x)h(x) = tan (x) = \frac{sin(x)}{cos(x)}

Solution: Although the derivative of ddx(tan(x))=sec2(x)\frac{d}{dx} (tan(x)) = sec^{2}(x) , but we will prove it using the quotient rule of how we arrived at that result.

Let's consider f(x)=sin(x)f(x) = sin(x) and g(x)=cos(x)g(x) = cos(x), then

h(x)=ddx(f(x)g(x))=f(x)g(x)g(x)f(x)(g(x))2h'(x)=\frac{d}{{dx}} \left(\frac{{f(x)}}{{g(x)}}\right) = \frac{{f'(x) \cdot g(x) - g'(x) \cdot f(x)}}{ ({g(x))^2}}
h(x)=(ddxsin(x))cos(x)(ddx(cos(x))sin(x)(cos(x))2h'(x) = \frac{(\frac{d}{dx}sin(x))\cdot cos(x) - (\frac{d}{dx}(cos(x)) \cdot sin(x)}{(cos(x))^{2}}
h(x)=cos(x)cos(x)(sin(x))sin(x)(cos(x))2h'(x) = \frac{cos(x) \cdot cos(x) - (-sin(x)) \cdot sin(x)}{(cos(x))^{2}}

Note: cos2(x)+sin2(x)=1 and 1cos(x)=sec(x)cos^2{(x)} + sin^{2}(x) = 1 \text{ and } \frac{1}{cos(x)}=sec(x)

h(x)=cos2(x)+sin2(x)cos2(x)=1cos2(x)=sec2(x)h'(x) = \frac{cos^2{(x)} + sin^{2}(x)}{cos^{2}(x)} = \frac{1}{cos^{2}(x)} = sec^{2}(x)

Hence, we have shown how using the quotient rule; we can find the derivative of tan(x)tan(x).

Quiz

Now that you know the quotient rule in differentiation, you can challenge yourself with a quiz.

Quotient rule in differentiation

Q

Differentiate the function h(x)=sin(x)x2h(x)= \frac{sin(x)}{x^{2}}

A)

h(x)=cos(x)(x2)sin(x)(2x)x2h'(x)= \frac{cos(x)(x^{2})-sin(x)(2x)}{x^{2}}

B)

h(x)=cos(x)(x2)sin(x)(2x)x4h'(x)= \frac{cos(x)(x^{2})-sin(x)(2x)}{x^{4}}

C)

h(x)=cos(x)2xh'(x)= \frac{cos(x)}{2x}

Conclusion

The quotient rule is an essential tool in Calculus for finding the derivative of a quotient of two functions. By applying the quotient rule correctly, we can efficiently compute derivatives and analyze the behavior of functions with fractions.

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