Quotient Rule Formula

Calculus is the mathematical study of continuous change, similar to how geometry is the study of shape and algebra is the study of arithmetic operations generalizations. Differential calculus and integral calculus are the two main disciplines; differential calculus is concerned with instantaneous rates of change and curve slopes, whereas integral calculus is concerned with the accumulation of quantities and areas under or between curves. The fundamental theorem of calculus connects these two disciplines, and they both employ the fundamental principles of infinite sequences and infinite series converge to a well-defined limit.

Quotient Rule Formula

In calculus, the quotient rule is a technique for determining the derivative of any function provided in the form of a quotient derived by dividing two differentiable functions. The quotient rule says that the derivative of a quotient is equal to the ratio of the result achieved by subtracting the numerator times the denominator’s derivative from the denominator times the numerator’s derivative to the square of the denominator’s derivative.

If we have a function of the type u(x)/v(x), we can use the quotient rule derivative to obtain the derivative of that function. The formula for the quotient rule is as follows:

where,

u(x) and v(x) are differentiable functions in R.

u'(x) and v'(x) are the derivatives of functions u(x) and v(x) respectively.

Derivation

Suppose a function f(x) = u(x)/v(x) is differentiable at x. We will prove the product rule formula using the definition of derivative or limits.

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Put  and 

= [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

This derives the formula for quotient rule.

Sample Problems

Question 1. Find the derivative of the function f(x) = 1/x using quotient rule.

Solution:

We have, f(x) = 1/x. Here, u(x) = 1 and v(x) = x.

So, u'(x) = 0 and v'(x) = 1

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [x (0) – 1 (1)/x²

= 1/x²

Question 2. Find the derivative of the function f(x) = 1/sin x using quotient rule.

Solution:

We have, f(x) = 1/sin x. Here, u(x) = 1 and v(x) = sin x.

So, u'(x) = 0 and v'(x) = cos x

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [sin x (0) – 1 (cos x)]/cos² x

= -cos x/ cos² x

= -1/cos x

= -sec x

Question 3. Find the derivative of the function f(x) = x/sin x using quotient rule.

Solution:

We have, f(x) = x/sin x. Here, u(x) = x and v(x) = sin x.

So, u'(x) = 1 and v'(x) = cos x

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [sin x (1) – x (cos x)]/cos² x

= (sin x – x cos x)/cos² x

Question 4. Find the derivative of the function f(x) = cos x/x using quotient rule.

Solution:

We have, f(x) = cos x/x. Here, u(x) = cos x and v(x) = x.

So, u'(x) = -sin x and v'(x) = 1

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [x (-sin x) – cos x (1)]/x²

= (-x sin x – cos x)/x²

Question 5. Find the derivative of the function f(x) = log x/x using quotient rule.

Solution:

We have, f(x) = log x/x. Here, u(x) = log x and v(x) = x.

So, u'(x) = 1/x and v'(x) = 1

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [x (1/x) – log x (1)]/x²

= (1 – log x)/x²

Question 6. Find the derivative of the function f(x) = (2x – 1)/x² using quotient rule.

Solution:

We have, f(x) = (2x – 1)/x². Here, u(x) = 2x – 1 and v(x) = x².

So, u'(x) = 2 and v'(x) = 2x

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [x² (2) – (2x – 1) (2x)]/x⁴

= (2x² – 4x² + 2x)/x⁴

= (-2x² + 2x)/x⁴

= [-2x(x – 1)]/x⁴

= -2(x – 1)/x³

Question 7. Find the derivative of the function f(x) = log x/sin x using quotient rule.

Solution:

We have, f(x) = log x/sin x. Here, u(x) = log x and v(x) = sin x.

So, u'(x) = 1/x and v'(x) = cos x

Using quotient rule we have,

f'(x) = [v(x) × u'(x) – u(x) × v'(x)]/[v(x)]²

= [sin x (1/x) – log x (cos x)]/sin² x

= [sin x/x – log x cos x]/sin² x