Let f: R rightarrow R be a continuous functio | vedclass

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(D) Given that $px+my+n=0$ is the tangent to the curve $y=f(x)$ at $x=\alpha$. The slope of the tangent line $px+my+n=0$ is $m_t = -\frac{p}{m}$. Ther

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$\frac{-2 p \alpha}{m}$

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(D) Given that $px+my+n=0$ is the tangent to the curve $y=f(x)$ at $x=\alpha$. The slope of the tangent line $px+my+n=0$ is $m_t = -\frac{p}{m}$. Therefore,$f'(\alpha) = -\frac{p}{m}$. Now,we need to find $\frac{d}{dx} [f(\alpha e^{2x})]$ at $x=0$. Using the chain rule: $\frac{d}{dx} [f(\alpha e^{2x})] = f'(\alpha e^{2x}) \cdot \frac{d}{dx}(\alpha e^{2x}) = f'(\alpha e^{2x}) \cdot (2\alpha e^{2x})$. At $x=0$,the expression becomes: $f'(\alpha e^0) \cdot (2\alpha e^0) = f'(\alpha) \cdot (2\alpha)$. Substituting $f'(\alpha) = -\frac{p}{m}$: $(-\frac{p}{m}) \cdot (2\alpha) = -\frac{2p\alpha}{m}$.

Let $f: R \rightarrow R$ be a continuous function. If $px+my+n=0$ is a tangent drawn to the curve $y=f(x)$ at $x=\alpha$,then at $x=0$,$\frac{d}{d x}\left(f\left(\alpha e^{2 x}\right)\right)=$

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