Differentiation Rules: Learn It 4

The Advanced Rules Cont.

The Quotient Rule

Having developed and practiced the product rule, we now consider differentiating quotients of functions.

As we see in the following theorem, the derivative of a quotient is not simply the quotient of the derivatives. Instead, it involves the derivative of the function in the numerator multiplied by the function in the denominator, minus the derivative of the function in the denominator multiplied by the function in the numerator, all divided by the square of the function in the denominator.

To better understand why we cannot just take the quotient of the derivatives, consider that:

[latex]\frac{d}{dx}(x^2)=2x[/latex],

which is not the same as,

[latex]\dfrac{\frac{d}{dx}(x^3)}{\frac{d}{dx}(x)} =\dfrac{3x^2}{1}=3x^2[/latex]

the quotient rule

Let [latex]f(x)[/latex] and [latex]g(x)[/latex] be differentiable functions. Then

[latex]\frac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)=\dfrac{\frac{d}{dx}(f(x))\cdot g(x)-\dfrac{d}{dx}(g(x))\cdot f(x)}{(g(x))^2}[/latex]

 

That is,

if [latex]j(x)=\dfrac{f(x)}{g(x)}[/latex], then [latex]j^{\prime}(x)=\dfrac{f^{\prime}(x)g(x)-g^{\prime}(x)f(x)}{(g(x))^2}[/latex]

The proof of the quotient rule is very similar to the proof of the product rule, so it is omitted here. Instead, we apply this new rule for finding derivatives in the next example.

Use the quotient rule to find the derivative of [latex]k(x)=\dfrac{5x^2}{4x+3}[/latex]

We explored the flexibility of the product rule given to the commutative property under addition and multiplication. It is worth mentioning that for the quotient rule, the order of the terms in the numerator will matter, as the commutative property does not hold under subtraction. We can see this from the example above:

[latex]{10x(4x+3)-4(5x^2)=20x^2+30x}[/latex]

however,

[latex]{4(5x^2)-10x(4x+3)=-20x^2-30x}[/latex].

Find the derivative of [latex]h(x)=\dfrac{3x+1}{4x-3}[/latex]


Extended Power Rule

It is now possible to use the quotient rule to extend the power rule to find derivatives of functions of the form [latex]x^k[/latex] where [latex]k[/latex] is a negative integer.

extended power rule

If [latex]k[/latex] is a negative integer, then

[latex]\dfrac{d}{dx}(x^k)=kx^{k-1}[/latex]

Proof

If [latex]k[/latex] is a negative integer, we may set [latex]n=−k[/latex], so that [latex]n[/latex] is a positive integer with [latex]k=−n[/latex]. Since for each positive integer [latex]n, \, x^{−n}=\frac{1}{x^n}[/latex], we may now apply the quotient rule by setting [latex]f(x)=1[/latex] and [latex]g(x)=x^n[/latex]. In this case, [latex]f^{\prime}(x)=0[/latex] and [latex]g^{\prime}(x)=nx^{n-1}[/latex]. Thus,

[latex]\frac{d}{dx}(x^{−n})=\dfrac{0(x^n)-1(nx^{n-1})}{(x^n)^2}[/latex].

 

Simplifying, we see that

[latex]\frac{d}{dx}(x^{−n})=\dfrac{−nx^{n-1}}{x^{2n}}=−nx^{(n-1)-2n}=−nx^{−n-1}[/latex].

 

Finally, observe that since [latex]k=−n[/latex], by substituting we have

[latex]\frac{d}{dx}(x^k)=kx^{k-1}[/latex]

[latex]_\blacksquare[/latex]

Find [latex]\frac{d}{dx}(x^{-4})[/latex]

Use the extended power rule and the constant multiple rule to find [latex]f(x)=\dfrac{6}{x^2}[/latex]