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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
Find $\frac{d y}{d x}$ if : $y=x^3-2 x^2+\sqrt{x}+1$
✓

Answer

$
y=x^3-2 x^2+\sqrt{x}+1
$
Differentiating w.r.t. $x$, we get
$
\begin{aligned}
\frac{\mathrm{d} y}{\mathrm{~d} x} & =\frac{\mathrm{d}}{\mathrm{d} x}\left(x^3-2 x^2+\sqrt{x}+1\right) \\
& =\frac{\mathrm{d}}{\mathrm{d} x}\left(x^3\right)-2 \frac{\mathrm{d}}{\mathrm{d} x}\left(x^2\right)+\frac{\mathrm{d}}{\mathrm{d} x}(\sqrt{x})+\frac{\mathrm{d}}{\mathrm{d} x}(1) \\
& =3 x^2-2(2 x)+\frac{\mathrm{d}}{\mathrm{d} x}\left(x^{\frac{1}{2}}\right)+0 \\
& =3 x^2-4 x+\frac{1}{2} x^{\frac{1}{2}-1} \text { } \\
& =3 x^2-4 x+\frac{1}{2} x^{\frac{-1}{2}} \\
\frac{\mathrm{d} y}{\mathrm{~d} x} & =3 x^2-4 x+\frac{1}{2 \sqrt{x}}
\end{aligned}
$

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