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2个回答
展开全部
y'=(2ˣ/x²)'
=[(2ˣ)'·x²-2ˣ·(x²)']/(x²)²
=[2ˣ·x²·ln2-2ˣ·2x]/x⁴
=(x·ln2-2)·2ˣ/x³
dy/dx|x=2
=(2·ln2-2)·2²/2³
=ln2-1
=[(2ˣ)'·x²-2ˣ·(x²)']/(x²)²
=[2ˣ·x²·ln2-2ˣ·2x]/x⁴
=(x·ln2-2)·2ˣ/x³
dy/dx|x=2
=(2·ln2-2)·2²/2³
=ln2-1
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