a. Chứng minh $\dfrac{y’}{cosx}-x=tanx$ với y=x.sinx b. Chứng minh $\dfrac{y’}{sinx}+x=cotx$ với y=x.cosx 08/09/2021 Bởi Ariana a. Chứng minh $\dfrac{y’}{cosx}-x=tanx$ với y=x.sinx b. Chứng minh $\dfrac{y’}{sinx}+x=cotx$ với y=x.cosx
\(\begin{array}{l}a)\quad y = x\sin x\\\to y’ = (x)’.\sin x + x.(\sin x)’\\\to y’ = \sin x + x\cos x\\\to \dfrac{y’}{\cos x} = \tan x + x\\\to \dfrac{y’}{\cos x} – x = \tan x\\b)\quad y = x\cos x\\\to y’ = (x)’.\cos x + x.(\cos x)’\\\to y’ = \cos x – x\sin x\\\to \dfrac{y’}{\sin x} = \cot x – x\\\to \dfrac{y’}{\sin x} + x = \cot x\end{array}\) Bình luận
\(\begin{array}{l}
a)\quad y = x\sin x\\
\to y’ = (x)’.\sin x + x.(\sin x)’\\
\to y’ = \sin x + x\cos x\\
\to \dfrac{y’}{\cos x} = \tan x + x\\
\to \dfrac{y’}{\cos x} – x = \tan x\\
b)\quad y = x\cos x\\
\to y’ = (x)’.\cos x + x.(\cos x)’\\
\to y’ = \cos x – x\sin x\\
\to \dfrac{y’}{\sin x} = \cot x – x\\
\to \dfrac{y’}{\sin x} + x = \cot x
\end{array}\)