# Differentiability

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# Let g(x) = f(x) sin x, where f(x) is a twice differentiable function on $\left(-\infty ,\infty \right)$ such that $\mathrm{f}\text{'}\left(-\mathrm{\pi }\right)=1$. The value of $\mathrm{g}\text{'}\text{'}\left(-\mathrm{\pi }\right)$ equals _____

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Solution

## We have g(x) = f(x) sin x                                                (1)On differentiating equation (1) w.r.t. x, we get$\mathrm{g}\text{'}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)\mathrm{cosx}+\mathrm{f}\text{'}\left(\mathrm{x}\right)\mathrm{sinx}$                                              (2)Again differentiating equation (2) w.r.t. x, we get $\mathrm{g}\text{'}\text{'}\left(\mathrm{x}\right)=\mathrm{f}\left(\mathrm{x}\right)\left(-\mathrm{sinx}\right)+\mathrm{f}\text{'}\left(\mathrm{x}\right)\mathrm{cosx}+\mathrm{f}\text{'}\left(\mathrm{x}\right)\mathrm{cosx}+\mathrm{f}\text{'}\text{'}\left(\mathrm{x}\right)\mathrm{sinx}$    (3)or $\mathrm{g}\text{'}\text{'}\left(-\mathrm{\pi }\right)=2\mathrm{f}\text{'}\left(-\mathrm{\pi }\right)\mathrm{cos}\left(-\mathrm{\pi }\right)=2×1×\left(-1\right)=-2$Hence, $\mathrm{g}\text{'}\text{'}\left(-\mathrm{\pi }\right)=-2$

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If function  is differentiable at $x=a$ then $\underset{x\to a}{lim} \frac{{x}^{2}f\left(a\right)-{a}^{2}f\left(x\right)}{x-a}$ is