2 replies to this topic

[25 minutes]

Tìm đạo hàm cấp n của $f(x)=\sqrt{ax+b}$

 

[200dong]

$y  = \sqrt{ax + b} = (ax + b)^{\frac{1}{2}}$

 

$y' = \frac{1}{2}(ax+b)^{\frac{1}{2} - 1} ; y'' = \frac{1}{2}(\frac{1}{2}-1)a^2(ax+b)^{\frac{1}{2} - 2}$

...

$y^{(n)} = \frac{1}{2}(\frac{1}{2} - 1)(\frac{1}{2} - n+1)a^n(ax +b)^{\frac{1}{2} -n}$

 

Tổng quát: 

 

$y =  \sqrt[k]{ax + b} = (ax + b)^{\frac{1}{k}}$

 

$y^{(n)} = \frac{1}{k}(\frac{1}{k} - 1)(\frac{1}{k} - n+1)a^n(ax +b)^{\frac{1}{k} -n}$

 

 

Edited by 200dong, 20-04-2013 - 18:24.

[25 minutes]

 

$y  = \sqrt{ax + b} = (ax + b)^{\frac{1}{2}}$

 

$y' = \frac{1}{2}(ax+b)^{\frac{1}{2} - 1} ; y'' = \frac{1}{2}(\frac{1}{2}-1)a^2(ax+b)^{\frac{1}{2} - 2}$

...

$y^{(n)} = \frac{1}{2}(\frac{1}{2} - 1)(\frac{1}{2} - n+1)a^n(ax +b)^{\frac{1}{2} -n}$

 

Tổng quát: 

 

$y =  \sqrt[k]{ax + b} = (ax + b)^{\frac{1}{k}}$

 

$y^{(n)} = \frac{1}{k}(\frac{1}{k} - 1)(\frac{1}{k} - n+1)a^n(ax +b)^{\frac{1}{k} -n}$

 

 

 

Theo tớ thì có thể viết lại thành $f^{(n)}(x)=\frac{(-1)^{n-1}a^n}{2^n}.(ax+b)^{\frac{1}{2}-n}$ được không ?