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Differentiation (p-1) — Maths (commerce) STD 11 Commerce / Arts — Question

Maharashtra BoardEnglish MediumSTD 11 Commerce / ArtsMaths (commerce)Differentiation (p-1)2 Marks
Question
Differentiate the following w.r.t. x. : $x \sqrt{x}+\log x-e^x$
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Answer

$
\begin{aligned}
& \text { Let } \mathrm{y}=\mathrm{x} \sqrt{ } \mathrm{x}+\log \mathrm{x}-\mathrm{e}^{\mathrm{x}} \\
& =x^{\frac{3}{2}}+\log x-e^x
\end{aligned}
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
& \frac{\mathrm{d} y}{\mathrm{~d} x}=\frac{\mathrm{d}}{\mathrm{d} x}\left(\begin{array}{c}
x^{\frac{3}{2}}+\log x-\mathrm{e}^x \\
\text { }
\end{array}\right. \\
& =\frac{\mathrm{d}}{\mathrm{d} x} x^{\frac{3}{2}}+\frac{\mathrm{d}}{\mathrm{d} x} \log x-\frac{\mathrm{d}}{\mathrm{d} x} \mathrm{e}^x \\
& =\frac{3}{2} x^{\frac{3}{2}-1}+\frac{1}{x}-\mathrm{e}^x \\
& =\frac{3}{2} x^{\frac{1}{2}}+\frac{1}{x}-\mathrm{e}^x \\
& =\frac{3}{2} \sqrt{x}+\frac{1}{x}-\mathrm{e}^x \\
&
\end{aligned}
$

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