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If $l_{1}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\sin \mathrm{x}}\right) \quad[2017-\mathrm{II}]$
$l_{2}=\lim _{h \rightarrow 0} \frac{e^{\sin (x+h)}-e^{\sin x}}{h}$
$l_{3}=\int \mathrm{e}^{\sin \mathrm{x}} \cos \mathrm{x} \mathrm{d} \mathrm{x}$
then which one of the following is correct?
Options:
$l_{2}=\lim _{h \rightarrow 0} \frac{e^{\sin (x+h)}-e^{\sin x}}{h}$
$l_{3}=\int \mathrm{e}^{\sin \mathrm{x}} \cos \mathrm{x} \mathrm{d} \mathrm{x}$
then which one of the following is correct?
Solution:
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Verified Answer
The correct answer is:
$\frac{\mathrm{d}}{\mathrm{dx}}\left(l_{3}\right)=l_{2}$
$l_{1}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\sin \mathrm{x}}\right)$
$l_{2}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{\sin (\mathrm{x}+\mathrm{h})}-\mathrm{e}^{\sin \mathrm{x}}}{\mathrm{h}}$
$l_{3}=\int \mathrm{e}^{\sin \mathrm{x}} \cdot \cos \mathrm{x} \mathrm{d} \mathrm{x}$
$l_{2}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{\sin (\mathrm{x}+\mathrm{h})}-\mathrm{e}^{\sin \mathrm{x}}}{\mathrm{h}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\sin \mathrm{x}}\right)=l_{1}$
$l_{3}=\int \mathrm{e}^{\sin \mathrm{x}} \cdot \cos \mathrm{x} \mathrm{d} \mathrm{x}$
Let $\sin \mathrm{x}=\mathrm{t} \Rightarrow \cos \mathrm{x} \mathrm{dx}=\mathrm{dt}$
$l_{3}=\int \mathrm{e}^{\mathrm{t}} \cdot \mathrm{dt}=\mathrm{e}^{\mathrm{t}}+\mathrm{c}=\mathrm{e}^{\sin \mathrm{x}}+\mathrm{c}$
$\frac{\mathrm{d}}{\mathrm{dx}}\left(l_{3}\right)=\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\mathrm{e}^{\sin \mathrm{x}}+\mathrm{c}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\sin \mathrm{x}}\right)=l_{1}=l_{2}$
$l_{2}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{\sin (\mathrm{x}+\mathrm{h})}-\mathrm{e}^{\sin \mathrm{x}}}{\mathrm{h}}$
$l_{3}=\int \mathrm{e}^{\sin \mathrm{x}} \cdot \cos \mathrm{x} \mathrm{d} \mathrm{x}$
$l_{2}=\lim _{\mathrm{h} \rightarrow 0} \frac{\mathrm{e}^{\sin (\mathrm{x}+\mathrm{h})}-\mathrm{e}^{\sin \mathrm{x}}}{\mathrm{h}}=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\sin \mathrm{x}}\right)=l_{1}$
$l_{3}=\int \mathrm{e}^{\sin \mathrm{x}} \cdot \cos \mathrm{x} \mathrm{d} \mathrm{x}$
Let $\sin \mathrm{x}=\mathrm{t} \Rightarrow \cos \mathrm{x} \mathrm{dx}=\mathrm{dt}$
$l_{3}=\int \mathrm{e}^{\mathrm{t}} \cdot \mathrm{dt}=\mathrm{e}^{\mathrm{t}}+\mathrm{c}=\mathrm{e}^{\sin \mathrm{x}}+\mathrm{c}$
$\frac{\mathrm{d}}{\mathrm{dx}}\left(l_{3}\right)=\frac{\mathrm{d}}{\mathrm{d} \mathrm{x}}\left(\mathrm{e}^{\sin \mathrm{x}}+\mathrm{c}\right)=\frac{\mathrm{d}}{\mathrm{dx}}\left(\mathrm{e}^{\sin \mathrm{x}}\right)=l_{1}=l_{2}$
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