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If \( y=f\left(x^{2}+2\right) \) and \( f^{\prime}(3)=5 . \) then \( \frac{d y}{d x} \) at \( x=1 \) is
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The correct answer is:
\( 10 \)
Given that, $y=\mathrm{f}\left(x^{2}+2\right) \rightarrow(1)$
and $\mathrm{f}^{\prime}(3)=5 \rightarrow(2)$
Now, $\frac{d y}{d x}=\mathrm{f}^{\prime}\left(x^{2}+2\right)(2 x)$
At $x=1$, we get $\left.\cdot \frac{d y}{d x}\right|_{x}=1=f^{\prime}(1+2)(2)=2 f(3)$
$=2(5)=10$
and $\mathrm{f}^{\prime}(3)=5 \rightarrow(2)$
Now, $\frac{d y}{d x}=\mathrm{f}^{\prime}\left(x^{2}+2\right)(2 x)$
At $x=1$, we get $\left.\cdot \frac{d y}{d x}\right|_{x}=1=f^{\prime}(1+2)(2)=2 f(3)$
$=2(5)=10$
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