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 $x$

$f{\bigl (}g(h(x)){\bigr )}$

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.

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

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

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

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

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

So we have

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

Thí dụ

• $f(x)=(x^{2}+1)^{3}$

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

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

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

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

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

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

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

• $f(x)=\sin(x^{2})$

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

$u(x)=x^{2}$

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

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

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

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

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

• $f(x)=|x|$

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

$u(x)=x^{2}$

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

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

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

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

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

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

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

• $f(x)=e^{\sin(x^{2})}=e^{g}$

$h(x)=x^{2}$

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

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

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

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

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

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