Google
Câu hỏi: Cho hàm số $f(x)=5 e^{x^2}$. Tính $P=f^{\prime}(x)-2 x . f(x)+\dfrac{1}{5} f(0)-f^{\prime}(0)$.
A. $P=4$.
B. $P=3$.
C. $P=2$.
D. $P=1$.
A. $P=4$.
B. $P=3$.
C. $P=2$.
D. $P=1$.
Ta có $f^{\prime}(x)=\left(5 e^{x^2}\right)^{\prime}=5 \cdot\left(x^2\right)^{\prime}\left(e^{x^2}\right)=10 x\left(e^{x^2}\right)$.
$
\begin{aligned}
& P=f^{\prime}(x)-2 x \cdot f(x)+\dfrac{1}{5} f(0)-f^{\prime}(0)=10 x\left(e^{x^2}\right)-2 x\left(5 e^{x^2}\right)+\dfrac{1}{5} \cdot\left(5 e^0\right)-10 \cdot 0 \cdot\left(e^0\right) \\
& =\dfrac{1}{5} \cdot 5=1 .
\end{aligned}
$
$
\begin{aligned}
& P=f^{\prime}(x)-2 x \cdot f(x)+\dfrac{1}{5} f(0)-f^{\prime}(0)=10 x\left(e^{x^2}\right)-2 x\left(5 e^{x^2}\right)+\dfrac{1}{5} \cdot\left(5 e^0\right)-10 \cdot 0 \cdot\left(e^0\right) \\
& =\dfrac{1}{5} \cdot 5=1 .
\end{aligned}
$
Đáp án D.