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If \(y=5 x^2+6 x+6, x=2, \Delta x=0.001\), then value of \(d y\) is
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Verified Answer
The correct answer is:
0.026
\(y=5 x^2+6 x+6, x=2\)
\(\begin{aligned}
\Delta x & =0.001 \\
d y & =(10 x+6) d x \\
d y & =(10 \times 2+6) \times 0.001 \\
d y & =0.026
\end{aligned}\)
\(\begin{aligned}
\Delta x & =0.001 \\
d y & =(10 x+6) d x \\
d y & =(10 \times 2+6) \times 0.001 \\
d y & =0.026
\end{aligned}\)
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