## Proving Quotient Rule Using First Principles

This property is called the differential rule of derivatives and is used to find the differentiation of the quotient of any two differential functions. Q.5: Write the quotient and the rest for the following: We can use the implicit differentiation method to derive the quotient rule for a differentiable function f(x) = u(x)/v(x), i.e. u(x) = f(x)⋅v(x). With the product rule, we have, u(x) = f(x)⋅v(x) + f(x)v(x). The resolution of f(x) gives that if 17 is divided by 3, we get the quotient 5 and the rest 2. (i.e.) 17 (dividend) ÷ 3 (divisor) = 5 (quotient) + 2 (remainder). Use the product limit rule to evaluate the product limit of two functions by evaluating the product of their limits. Using the product rule f(x) = u(x)v-1(x) + u(x)⋅(left(frac{d}{dx}(v^{-1}(x)right)) ⇒f(x) = u(x)v-1(x) + u(x)⋅(-1)(v(x)-2)v(x) ⇒f(x) = (frac{d}{dx} left[ frac{u(x)}} {v(x)} right] ) = (frac{u(x)v(x) – u(x)v(x)}{[v(x)]^2}) The quotient rule of derivatives is $$frac{d}{{dx}}left[ {frac{{fleft( x right)}}{{ g left( x right)}}} right] = frac{{gleft( x right)fleft( x right) – fleft( x right)gleft( x right)}}{{{left[ {gleft( x right)} right]}^2}}}$$$. In other words, we can interpret this to mean that the derivative of a quotient of two functions is equal to the second function as it is, and the derivative of the first function minus the first function as it is, and the derivative of the second function divided by the square of the second function. This rule can be proved by the first principle or the derivative by definition. Now consider the other example, 15 ÷ 2. In this case, 15 is not exactly divisible by 2, so we get the quotient value 7 and the rest 1. Yes, the quotient should be the solution to any division problem. If a number 5 is divided by 5, we get the quotient 1, which is the answer to the division problem. When we work on issues of division, we must first determine what dividend and which divisor are. To obtain the quotient of a number, the dividend is divided by the divisor. This means that the problem should be in the following form: In mathematics, a quotient is a resulting number when one number is divided by the other number. In other words, a dividend is divided by a divisor, we get the profit quotient. The quotient is commonly used in mathematics and is usually referred to as a fraction or ratio. Lets see its representation to better understand it. ⇒ f(x) = u(x)v-1(x) Using the product rule f(x) = u(x)v-1(x) + u(x)⋅(left(frac{d}{dx}(v^{-1}(x)right)) In this case, dividend 12 is perfectly divided by 2. So we get the quotient value 6 and the rest 0. If we apply the power rule to solve the derivative in the second term, we have, f(x) = u(x)v-1(x) + u(x)⋅(-1)(v(x)-2)v(x) In the above representation, we can see that the numerator is the dividend and the denominator of the divisor. If the numerator is divided by the denominator, the result is the quotient. Now lets look at the representation in terms of the long division method in the figure below. In the numerator,$g{(x)}$is a common factor in the first two terms and$f{(x)}$is a common factor in the remaining two terms. So take them together to take a first step towards simplifying this mathematical expression. In mathematics, the quotient is the number generated when we perform division operations on two numbers. Basically, this is the result of the method of division. There are four main terminologies used in arithmetic division, such as divisor, dividend, quotient, and rest. Each term is explained here in this article with meaning and examples. Also learn how to share numbers here Evaluate the limit of the first factor of each term in the first factor and the second factor using the direct substitution method. The quotient formula is: Dividend ÷ divisor = quotient (if the remainder is zero) The general formula for each division problem is given by: dividend ÷ divisor = quotient + remainder. by factoring #g(x)# of the first two terms and #-f(x)# of the last two terms, With the quotient rule of a derivative, we now have [begin{gathered}frac{{dy}}{{dx}} = frac{{left( {{x^3} + 8} right)frac{d}{{dx}}left( {{x^3} – 8} right) – left( {{x^3} – 8} right)frac{d}{{dx}}left( {{x^3} + 8} right)}}{{left( {{x^3} + 8} right)}^2}}} \frac{{dy}}{{dx}} = frac{{left( {{x^3} + 8} right)3{x^ 2} – left( {{x^3} – 8} right)3{x^2}}}{{left( {{x^3} + 8} right)}^2}}} frac{{dy}}{{dx}} = frac{{left( {3{x^5} + 24{x^2}} right) – left( {3{x^5} – 24{x^2}} right)}} }{{{ left( {{x^3} + 8} right)}^2}}} frac{{dy}}{{dx}} = frac{{48{x^2}}}{{{left( {{x^3} + 8} right)}^2}}} end{gathered} ] According to the definition of the derivative The derivative of the quotient of two differential functions can be written as a limit operation, find the differentiation of the quotient according to the first principle. When a dividend is divided by a divisor, we get the result. So, basically, division is the inverse process of multiplication, so if we multiply the quotient and the divisor, we get the dividend. Thus, we can also define the divisor, dividend and the rest here. The differentiation quotient rule is written in two different forms, taking$u = f{(x)}$and$v = g{(x)}\$. In the division method, a number is divided by another number to obtain another output number. Here, the number/integer that is divided is called the dividend, and the integer that divides a given number is the divisor. The divisor that does not completely divide a number gives a number called the rest. The separator is marked with ÷ or /`.