Đạo hàm Giúp e câu 3 với câu 7 Question Đạo hàm Giúp e câu 3 với câu 7 in progress 0 Môn Toán Nem 5 years 2021-05-16T01:51:40+00:00 2021-05-16T01:51:40+00:00 2 Answers 27 views 0
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Đáp án:
Giải thích các bước giải:
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Đáp án:
\(\begin{array}{l}
3)\quad y’ = \dfrac{x^2 + 4x – 5}{(2+x)^2}\\
7)\quad y’ = -2\cos^2\left(\dfrac{2x+1}{3}\right).\sin\left(\dfrac{2x+1}{3}\right)\\
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
3)\quad y = \dfrac{x^2 – 2x+1}{2+x}\\
\to y’ = \dfrac{(x^2 – 2x + 1)'(2+x) – (x^2 – 2x + 1)(2+x)’}{(2+x)^2}\\
\to y’ = \dfrac{(2x-2)(2+x) – (x^2 – 2x + 1)}{(2+x)^2}\\
\to y’ = \dfrac{x^2 + 4x – 5}{(2+x)^2}\\
7)\quad y = \cos^3\left(\dfrac{2x+1}{3}\right)\\
\to y’ = 3\cos^2\left(\dfrac{2x+1}{3}\right).\left[\cos\left(\dfrac{2x+1}{3}\right)\right]’\\
\to y’ = 3\cos^2\left(\dfrac{2x+1}{3}\right)\cdot \left[-\left(\dfrac{2x+1}{3}\right)\right]’.\sin\left(\dfrac{2x+1}{3}\right)\\
\to y’ = 3\cos^2\left(\dfrac{2x+1}{3}\right)\cdot\left(-\dfrac{2}{3}\right).\sin\left(\dfrac{2x+1}{3}\right)\\
\to y’ = -2\cos^2\left(\dfrac{2x+1}{3}\right).\sin\left(\dfrac{2x+1}{3}\right)\\
\end{array}\)