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68. If $f(x)=\frac{e^{-x} \sin x}{\log _e x}$ and $f^{\prime}(x)=f(x) . g(x)$, then $g^{\prime}(\mathrm{e})=$
(a) $e^{-2}-\operatorname{cosec}^2(e)$
(b) $2 e^{-2}-\operatorname{cosec}^2(e)$
(c) $2 e^{-2}-\operatorname{cosec}^2(e)$
(d) $2 e^{-2}+\operatorname{cosec}^2(e)$
Options:
(a) $e^{-2}-\operatorname{cosec}^2(e)$
(b) $2 e^{-2}-\operatorname{cosec}^2(e)$
(c) $2 e^{-2}-\operatorname{cosec}^2(e)$
(d) $2 e^{-2}+\operatorname{cosec}^2(e)$
Solution:
2064 Upvotes
Verified Answer
The correct answer is:
$2 e^{-2}-\operatorname{cosec}^2(e)$
(c) Given $\mathrm{f}(\mathrm{x})=\frac{\mathrm{e}^{-\mathrm{x}} \sin \mathrm{x}}{\log _{\mathrm{e}}^{\mathrm{x}}}$
Differentiate w.r.t. ' $x$ ' both sides
$$
f^{\prime}(x)=\frac{\left(e^{-x} \cos x-e^{-x} \sin x\right) \log ^x-e^{-x} \frac{\sin x}{x}}{\left(\log e^x\right)^2}
$$
Now take $\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})$
$$
\frac{e^{-x}(\cos x-\sin x) \log e^x}{\left(\log e^x\right)^2}-\frac{e^{-x} \sin x}{x\left(\log e^x\right)^2}=\frac{e^{-x} \sin x}{\log e^x} g(x)
$$
Divide bothside by $\frac{\mathrm{e}^{-\mathrm{x}} \sin \mathrm{x}}{\log \mathrm{e}^{\mathrm{x}}}$
$$
\begin{aligned}
& g(x)=\frac{\frac{e^{-x}(\cos x-\sin x)}{\left(\log e^x\right)}}{\frac{e^{-x} \sin x}{\left(\log e^x\right)}}-\frac{\frac{e^{-x} \sin x}{x\left(\log e^x\right)^2}}{\frac{e^{-x} \sin x}{\log e^x}} \\
& g(x)=\frac{\cos x}{\sin x}-\frac{1}{x \log e^x} \\
& g(x)=\cot x-1-\frac{1}{x \log e^x}
\end{aligned}
$$
7
Differentiate w.r.t. ' $x$ ' both sides
$$
g^{\prime}(x)=-\operatorname{cosec}^2 x+1-\frac{\log _e^x+1}{x^2\left(\log _e^x\right)^2}
$$
put $\mathrm{x}=\mathrm{e}$ in $\mathrm{g}^{\prime}(\mathrm{x})$
$$
\begin{aligned}
& \mathrm{g}^{\prime}(\mathrm{e})=-\operatorname{cosec}^2(\mathrm{e})+\frac{\left(\log _{\mathrm{e}^{e+1}}\right)}{\left(\mathrm{e}^2\right)\left(\log _{\mathrm{e}^e}\right)^2} \\
& \mathrm{~g}^{\prime}(\mathrm{e})=-\operatorname{cosec}^2(\mathrm{e})+\frac{2}{\mathrm{e}^2 1}=2 \mathrm{e}^2-\operatorname{cosec}^2(\mathrm{e})
\end{aligned}
$$
so, option (c) is correct
Differentiate w.r.t. ' $x$ ' both sides
$$
f^{\prime}(x)=\frac{\left(e^{-x} \cos x-e^{-x} \sin x\right) \log ^x-e^{-x} \frac{\sin x}{x}}{\left(\log e^x\right)^2}
$$
Now take $\mathrm{f}^{\prime}(\mathrm{x})=\mathrm{f}(\mathrm{x}) \mathrm{g}(\mathrm{x})$
$$
\frac{e^{-x}(\cos x-\sin x) \log e^x}{\left(\log e^x\right)^2}-\frac{e^{-x} \sin x}{x\left(\log e^x\right)^2}=\frac{e^{-x} \sin x}{\log e^x} g(x)
$$
Divide bothside by $\frac{\mathrm{e}^{-\mathrm{x}} \sin \mathrm{x}}{\log \mathrm{e}^{\mathrm{x}}}$
$$
\begin{aligned}
& g(x)=\frac{\frac{e^{-x}(\cos x-\sin x)}{\left(\log e^x\right)}}{\frac{e^{-x} \sin x}{\left(\log e^x\right)}}-\frac{\frac{e^{-x} \sin x}{x\left(\log e^x\right)^2}}{\frac{e^{-x} \sin x}{\log e^x}} \\
& g(x)=\frac{\cos x}{\sin x}-\frac{1}{x \log e^x} \\
& g(x)=\cot x-1-\frac{1}{x \log e^x}
\end{aligned}
$$
7
Differentiate w.r.t. ' $x$ ' both sides
$$
g^{\prime}(x)=-\operatorname{cosec}^2 x+1-\frac{\log _e^x+1}{x^2\left(\log _e^x\right)^2}
$$
put $\mathrm{x}=\mathrm{e}$ in $\mathrm{g}^{\prime}(\mathrm{x})$
$$
\begin{aligned}
& \mathrm{g}^{\prime}(\mathrm{e})=-\operatorname{cosec}^2(\mathrm{e})+\frac{\left(\log _{\mathrm{e}^{e+1}}\right)}{\left(\mathrm{e}^2\right)\left(\log _{\mathrm{e}^e}\right)^2} \\
& \mathrm{~g}^{\prime}(\mathrm{e})=-\operatorname{cosec}^2(\mathrm{e})+\frac{2}{\mathrm{e}^2 1}=2 \mathrm{e}^2-\operatorname{cosec}^2(\mathrm{e})
\end{aligned}
$$
so, option (c) is correct
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