3 years ago

Write 19.4 in expanded form

Mathematics

1 answer:

Natali5045456 [20]3 years ago

6 0

The answer is 10+9+0.4

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$\huge\boxed{f'(x)=-\dfrac{e^\frac{1}{x}}{x^2}}$

Step-by-step explanation:

$(e^x)'=e^x\\\\\bigg[f\bigg(g(x)\bigg)\bigg]'=f'\bigg(g(x)\bigg)\cdot g'(x)\\\\\left(\dfrac{1}{x}\right)'=-\dfrac{1}{x^2}\\\\====================$

$f(x)=e^\frac{1}{x}=\left(e^\frac{1}{x}\right)'\cdot\left(\dfrac{1}{x}\right)'=e^\frac{1}{x}\cdot\left(-\dfrac{1}{x^2}\right)=-\dfrac{e^\frac{1}{x}}{x^2}$

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Step-by-step explanation:

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