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Serjik [45]
3 years ago
10

I

Mathematics
1 answer:
Liula [17]3 years ago
8 0

Answer:

2-6

Step-by-step explanation:

5 673226

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Ostrovityanka [42]

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Step-by-step explanation:

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Gnesinka [82]
Given:
f(x)=4xcos^{-1}(2x+4)-\sqrt{3-3x^2}

Using
\frac{d}{dx}cos^{-1}(x)=-\frac{1}{\sqrt{1-x^2}}
we derive
\frac{d}{dx}4xcos^{-1}(2x+4)
=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}

Similarly, using
\frac{d}{dx}\sqrt{x}=\frac{1}{2\sqrt{x}}
we derive
\frac{d}{dx}(-\sqrt{3-3x^2})
=\frac{3x}{\sqrt{3-3x^2}}

Therefore, the derivative is
f'(x)=\frac{d}{dx}(4xcos^{-1}(2x+4)-\sqrt{3-3x^2})
=4cos^{-1}(2x+4)-\frac{8x}{\sqrt{1-(2x+4)^2}}+\frac{3x}{\sqrt{3-3x^2}}
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