Sách giải tích/Đạo hàm/Đạo hàm hàm số kép

Tìm đạo hàm của hàm số trong hàm số

Công thức

Nếu có hàm số $f$ là hàm số của $g$ và là hàm số của $h$ và là hàm số của $x$

$f(g(h(x)))$

Thì đạo hàm của hàm số $f$ với biến số $x$ sẻ là

$\frac{df}{dx}=\frac{df}{dg}\cdot \frac{dg}{dh}\cdot \frac{dh}{dx}$

$\frac{df}{dx}=\frac{df}{\cancel{dg}}\cdot \frac{\cancel{dg}}{\cancel{dh}}\cdot \frac{\cancel{dh}}{dx}$

Chứng minh

Suppose $y$ is a function of $u$ which is a function of $x$ (it is assumed that $y$ is differentiable at $u$ and $x$ , and $u$ is differentiable at $x$ . To prove the chain rule we use the definition of the derivative.

$\frac{dy}{dx}=\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}$

We now multiply $\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}$ by $\frac{\mathrm{\Delta }u}{\mathrm{\Delta }u}$ and perform some algebraic manipulation.

$\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\mathrm{\Delta }y}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\mathrm{\Delta }y}{\mathrm{\Delta }u}\cdot \frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\mathrm{\Delta }y}{\mathrm{\Delta }u}\cdot \underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\mathrm{\Delta }u}{\mathrm{\Delta }x}=\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\mathrm{\Delta }y}{\mathrm{\Delta }u}\cdot \frac{du}{dx}$

Note that as $\mathrm{\Delta }x$ approaches $0$ , $\mathrm{\Delta }u$ also approaches $0$ . So taking the limit as of a function as $\mathrm{\Delta }x$ approaches $0$ is the same as taking its limit as $\mathrm{\Delta }u$ approaches $0$ . Thus

$\underset{\mathrm{\Delta }x\to 0}{\lim }\frac{\mathrm{\Delta }y}{\mathrm{\Delta }u}=\underset{\mathrm{\Delta }u\to 0}{\lim }\frac{\mathrm{\Delta }y}{\mathrm{\Delta }u}=\frac{dy}{du}$

So we have

$\frac{dy}{dx}=\frac{dy}{du}\cdot \frac{du}{dx}$

Thí dụ

• $f(x)=(x^2+1{)}^3$

$u(x)=x^2+1$ |

$f(x)=u(x{)}^3$

$\frac{df}{dx}=\frac{df}{du}\cdot \frac{du}{dx}$

$\frac{df}{dx}=\frac{d}{du}{u}^3\cdot \frac{d}{dx}(x^2+1)$

$\frac{df}{dx}=3u^2\cdot 2x$

$\frac{df}{dx}=3(x^2+1{)}^2\cdot 2x$

$\frac{df}{dx}=6x(x^2+1{)}^2$

• $f(x)=\sin (x^2)$

$f(x)=\sin (x^2)$

$u(x)=x^2$

$f(x)=\sin (u)$

$\frac{df}{dx}=\frac{df}{du}\cdot \frac{du}{dx}$

$\frac{df}{dx}=\frac{d}{du}\sin (u)\cdot \frac{d}{dx}(x^2)$

$\frac{df}{dx}=\cos (u)\cdot 2x$

$\frac{df}{dx}=\cos (x^2)\cdot 2x$

• $f(x)=|x|$

$f(x)=\sqrt{x^2}$

$u(x)=x^2$

$f(x)=u(x{)}^{\frac{1}{2}}$

$\frac{df}{dx}=\frac{df}{du}\cdot \frac{du}{dx}$

$\frac{df}{dx}=\frac{d}{du}{u}^{\frac{1}{2}}\cdot \frac{d}{dx}(x^2)$

$\frac{df}{dx}=\frac{u^{-\frac{1}{2}}}{2}\cdot 2x$

$\frac{df}{dx}=\frac{(x^2{)}^{-\frac{1}{2}}}{2}\cdot 2x$

$\frac{df}{dx}=\frac{x}{\sqrt{x^2}}$

$\frac{df}{dx}=\frac{x}{|x|}$

• $f(x)=e^{\sin (x^2)}=e^g$

$h(x)=x^2$

$g(x)=\sin (h)=\sin (x^2)$

$\frac{df}{dx}=\frac{df}{dg}\cdot \frac{dg}{dh}\cdot \frac{dh}{dx}$

$\frac{df}{dg}=e^g=e^{\sin (x^2)}$

$\frac{dg}{dh}=\cos (h)=\cos (x^2)$

$\frac{dh}{dx}=2x$

$\frac{d}{dx}{e}^{\sin (x^2)}=e^{\sin (x^2)}\cdot \cos (x^2)\cdot 2x$