Find the decreasing range of the function:
$$f(x)=xe^{\sin x}.$$
I found derivative $$f'(x)=(xe^{\sin x})'=e^{\sin x} (x \cos x + 1)$$
But I couldn't solve this inequality:
$$f'(x)≤0 \Longrightarrow e^{\sin x} (x \cos x + 1)≤0$$
$$x \cos x + 1≤0$$
Find the decreasing range of the function:
$$f(x)=xe^{\sin x}.$$
I found derivative $$f'(x)=(xe^{\sin x})'=e^{\sin x} (x \cos x + 1)$$
But I couldn't solve this inequality:
$$f'(x)≤0 \Longrightarrow e^{\sin x} (x \cos x + 1)≤0$$
$$x \cos x + 1≤0$$
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