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Quotient Rule

According to the Quotient Rule, a quotient's derivative equals the denominator times the numerator's derivative minus the numerator times the denominator's derivative, all divided by the denominator's square.

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

That is, if we need to compute the derivative of a function of the type f(x)/g(x), we can use the quotient technique as long as both f(x) and g(x) are differentiable and g(x) ≠0. The quotient rule directly follows the product rule and the concept of derivation limits in differentiation.

Quotient rule formula

The quotient rule is a technique for calculating the derivative of a function that is the ratio of two differentiable functions in calculus.

The quotient rule derivative formula can be used to calculate the derivative or evaluate the differentiation of a quotient of two functions. The derivative formula for the quotient rule is as follows:

Let,fx=g(x)h(x), g and h are both differentiable, and h(x)≠0 According to the quotient rule, the derivative of f(x) equals

fx=g’xhx-gxh'(x)h(x)2

Quotient Rule Formula Derivation

We learned how to get the derivatives of the quotient of two differentiable functions using the quotient formula in the previous section. The quotient rule formula is proved. There are several ways to prove the quotient rule formula, including,

  • Using derivative and limit properties

  • Using implicit differentiation

  • Using chain rule

Proof from derivative & limit properties

If fx=g(x)h(x) 

The following proof is obtained by using the derivative definition and the properties of limits.

f’x=fx+k-f(x)k

g(x+k)h(x+k)-g(x)h(x)k

gx+khx-gxh(x+k)k.hxh(x+k)

gx+kh(x)-gxh(x+k)k .k→01hxh(x+k)

gx+khx-gxhxgxhx-gxh(x+k)k ).1h(x)2

gx+khx-gxhxk -gxgx+k-gxhxk ).1h(x)2

gx+k-gxk .g(x)hx+k-hxk ).1h(x)2

=gxhx-gxh'(x)h(x)2

Using implicit differentiation

Let,  fx=g(x)f(x), , so gx=fxhxThe product rule is then appliedg’x=f’xhxfxh'(x)   Problem-solving for f'(x)  and substituting for fx  gives:             

f’x=g’x-fxh'(x)h(x)

=g’x-gxhx.h'(x)h(x)

=g’xhx-gxh'(x)h(x)2

Using chain rule

Let fx=g(x)h(x)=gxh(x)-1  Then there is the product rule.

f’x=g’xh(x)-1+gx.ddx(h(x)-1)

Apply the power rule and the chain rule to evaluate the derivative in the second term:

f’x=g’xh(x)-1+gx.-1h(x)-2h'(x)

Finally, rewrite the equation as fractions and combine terms to arrive at

f’x=g'(x)h(x)-gxh'(x)h(x)2

=g’xhx-gxh'(x)h(x)2

How do you do the quotient rule

Step are given below:

  • Make the expression easier to understand.

  • Use the product rule to help you.

  • The derivative of each part is calculated.

  • Simplify by substituting derivatives into the product rule.

Example: 1

fx=x31-x4 , find f'(x)

Solution:

f’x=1-x4ddxx3-x3ddx(1-x4)(1-x4)2

=1-x43x2-(x3)(-4×3)(1-x4)2

=3×2-3×6+4×6(1-x4)2

=3×2+x6(1-x4)2

Example: 2 Determine the function’s derivative: fx=x-1x+2.

Solution:

Because this is a fraction involving two functions, we use the quotient rule first.

f’x=x-1’x+2-(x-1)(x+2)'(x+2)

For each prime, get the derivative.

f’x=(1)x+2-(x-1)(1)(x+2)2

f’x=x+2-(x-1)(x+2)2

Simplify it,

=x+2-x+1(x+2)2

=3(x+2)2

Conclusion

The quotient is the result of dividing two numbers. Six divided by two equals three as a quotient. The word “quotient” from Latin for “how many times.” That makes sense: dividing one number by another determines “how many times” the second number enters the first.  And we learnt how to utilize the quotient formula to find the derivatives of the quotient of two differentiable functions. Let’s have a look at how the quotient rule formula is proved. The quotient rule formula can be proved in a variety of methods, including: Using derivative and limit properties, using implicit differentiation, using chain rule

It is critical for us to understand how to estimate because we may not always require an exact solution to an issue. Estimation is also a great technique for pupils to double-check their quotients to make sure they’re correct.

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