Sách giải tích/Đạo hàm/Đạo hàm tích 2 hàm số

Công thức

$\frac{d}{dx}[f(x)\cdot g(x)]=f(x)\cdot g^{\prime }(x)+f^{\prime }(x)\cdot g(x)$

$(f\cdot g{)}^{\prime }=f^{\prime }\cdot g+f\cdot g^{\prime }$

Chứng minh

Proving this rule is relatively straightforward, first let us state the equation for the derivative:

$\frac{d}{dx}[f(x)\cdot g(x)]=\underset{h\to 0}{\lim }\frac{f(x+h)\cdot g(x+h)-f(x)\cdot g(x)}{h}$

We will then apply one of the oldest tricks in the book—adding a term that cancels itself out to the middle:

$\frac{d}{dx}[f(x)\cdot g(x)]=\underset{h\to 0}{\lim }\frac{f(x+h)\cdot g(x+h)\mathbf{-}𝐟\mathbf{(}𝐱\mathbf{+}𝐡\mathbf{)}\mathbf{\cdot }𝐠\mathbf{(}𝐱\mathbf{)}\mathbf{+}𝐟\mathbf{(}𝐱\mathbf{+}𝐡\mathbf{)}\mathbf{\cdot }𝐠\mathbf{(}𝐱\mathbf{)}-f(x)\cdot g(x)}{h}$

Notice that those terms sum to zero, and so all we have done is add 0 to the equation. Now we can split the equation up into forms that we already know how to solve:

$\frac{d}{dx}[f(x)\cdot g(x)]=\underset{h\to 0}{\lim }[\frac{f(x+h)\cdot g(x+h)-f(x+h)\cdot g(x)}{h}+\frac{f(x+h)\cdot g(x)-f(x)\cdot g(x)}{h}]$

Looking at this, we see that we can separate the common terms out of the numerators to get:

$\frac{d}{dx}[f(x)\cdot g(x)]=\underset{h\to 0}{\lim }[f(x+h)\frac{g(x+h)-g(x)}{h}+g(x)\frac{f(x+h)-f(x)}{h}]$

Which, when we take the limit, becomes:

$\frac{d}{dx}[f(x)\cdot g(x)]=f(x)\cdot g^{\prime }(x)+g(x)\cdot f^{\prime }(x)$, or the mnemonic "one D-two plus two D-one"

This can be extended to 3 functions:

$\frac{d}{dx}[fgh]=f(x)g(x)h^{\prime }(x)+f(x)g^{\prime }(x)h(x)+f^{\prime }(x)g(x)h(x)$

For any number of functions, the derivative of their product is the sum, for each function, of its derivative times each other function.

Back to our original example of a product, $h(x)=(x^2+5x+7)\cdot (x^3+2x-4)$, we find the derivative by the product rule is

$h^{\prime }(x)=(x^2+5x+7)(3x^2+2)+(2x+5)(x^3+2x-4)=5x^4+20x^3+27x^2+12x-6$

Note, its derivative would not be

$(2x+5)\cdot (3x^2+2)=3x^3+15x^2+4x+10$

which is what you would get if you assumed the derivative of a product is the product of the derivatives.

To apply the product rule we multiply the first function by the derivative of the second and add to that the derivative of first function multiply by the second function. Sometimes it helps to remember the memorize the phrase "First times the derivative of the second plus the second times the derivative of the first."