Differentiate the following expression w.r.t. $x$
$\sqrt{x}\left( {{x}^{3}}+{{x}^{2}}-3x \right)$
(a) $\dfrac{1}{2\sqrt{x}}\left( 7{{x}^{3}}+5{{x}^{2}}+9x \right)$
(b) $\dfrac{1}{2\sqrt{x}}\left( 7{{x}^{3}}+5{{x}^{2}}-9x \right)$
(c) $\dfrac{1}{2\sqrt{x}}\left( 7{{x}^{3}}-5{{x}^{2}}-9x \right)$
(d) $\dfrac{1}{2\sqrt{x}}\left( -7{{x}^{3}}+5{{x}^{2}}-9x \right)$

Answer Verified Verified
Hint: Use multiplication rule of differentiation. It is given as
$\dfrac{d}{dx}\left( u.v \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}$
Apply this formula to get the differentiation of the equation given. Use the relation
$\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ to get the answer.

Complete step-by-step solution -
As we need to find the differentiation of $\sqrt{x}\left( {{x}^{3}}+{{x}^{2}}-3x \right)$ . So, let us suppose the given expression is represented by $'y'$ .So, we get
$y=\sqrt{x}\left( {{x}^{3}}+{{x}^{2}}-3x \right)........\left( i \right)$
As we can observe the equation(i) and get that $\sqrt{x}$ and ${{x}^{3}}+{{x}^{2}}-3x$ are in multiplication form. So, we can use multiplication rule of differentiation, which is given as
$\dfrac{d}{dx}\left( uv \right)=u\dfrac{dv}{dx}+v\dfrac{du}{dx}..........\left( ii \right)$
Now, we can put value of ‘u’ as $\sqrt{x}$ and ‘v’ as ${{x}^{2}}+{{x}^{2}}-3x$ and use the relation given in equation(i). So, we get
$\dfrac{d}{dx}\left( \sqrt{x}\left( {{x}^{3}}+{{x}^{2}}-3x \right) \right)=\dfrac{dy}{dx}=\sqrt{x}\dfrac{d}{dx}\left( {{x}^{3}}+{{x}^{2}}-3x \right)+\left( {{x}^{3}}+{{x}^{2}}-3x \right)\dfrac{d}{dx}\left( \sqrt{x} \right)$
Now, we know the derivative of ${{x}^{n}}$ is given as
$\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}..........\left( iii \right)$
Hence, we can get value of $\dfrac{dy}{dx}$ as
$\dfrac{dy}{dx}={{x}^{\dfrac{1}{2}}}\left[ \dfrac{d}{dx}{{x}^{3}}+\dfrac{d}{dx}{{x}^{2}}-3\dfrac{d}{dx}x \right]+\left( {{x}^{3}}+{{x}^{2}}-3x \right)\dfrac{d}{dx}\left( {{x}^{\dfrac{1}{2}}} \right)$
Now using the relation (iii), we get
$\dfrac{dy}{dx}={{x}^{\dfrac{1}{2}}}\left[ 3{{x}^{2}}+2x-3 \right]+\left( {{x}^{3}}+{{x}^{2}}-3x \right)\times \dfrac{1}{2}{{x}^{-\dfrac{1}{2}}}$
$\dfrac{dy}{dx}={{x}^{\dfrac{1}{2}}}\left[ 3{{x}^{2}}+2x-3 \right]+\left( {{x}^{3}}+{{x}^{2}}-3x \right)\times \dfrac{{{x}^{-\dfrac{1}{2}}}}{2}$
Now, we know the property of surds given as
  & {{a}^{-m}}=\dfrac{1}{{{a}^{m}}}.........\left( iv \right) \\
 & {{a}^{m}}.{{a}^{n}}={{a}^{m+n}}.....\left( v \right) \\
Now, we can use above relations and hence get the value of $\dfrac{dy}{dx}$ as
  & \dfrac{dy}{dx}={{x}^{\dfrac{1}{2}}}\left[ 3{{x}^{2}}+2x-3 \right]+\left( {{x}^{3}}+{{x}^{2}}-3x \right)\times \dfrac{1}{2{{x}^{\dfrac{1}{2}}}} \\
 & \dfrac{dy}{dx}={{x}^{\dfrac{1}{2}}}\left[ 3{{x}^{2}}+2x-3 \right]+\dfrac{\left( {{x}^{3}}+{{x}^{2}}-3x \right)}{2{{x}^{\dfrac{1}{2}}}} \\
 & \dfrac{dy}{dx}=\dfrac{2{{x}^{\dfrac{1}{2}}}{{x}^{\dfrac{1}{2}}}\left( 3{{x}^{2}}+2x-3 \right)+\left( {{x}^{3}}+{{x}^{2}}-3x \right)}{2{{x}^{\dfrac{1}{2}}}} \\
 & \\
 & \dfrac{dy}{dx}=\dfrac{2x\left( 3{{x}^{2}}+2x-3 \right)+\left( {{x}^{3}}+{{x}^{2}}-3x \right)}{2{{x}^{\dfrac{1}{2}}}} \\
 & \dfrac{dy}{dx}=\dfrac{6{{x}^{3}}+4{{x}^{2}}-6x+{{x}^{3}}+{{x}^{2}}-3x}{2\sqrt{x}} \\
 & \dfrac{dy}{dx}=\dfrac{7{{x}^{3}}+5{{x}^{2}}-9x}{2\sqrt{x}} \\
 & \Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{2}x}\left( 7{{x}^{3}}+5{{x}^{2}}-9x \right) \\
Hence, differentiation of the given expression in the problem is given by equation.
So, option(b) is the correct answer.

Note: Another approach for the given problem would be that we can multiply the term $\sqrt{x}$ to the terms of the bracket. And hence, using $\dfrac{d}{dx}{{x}^{n}}=n{{x}^{n-1}}$ only, we can solve the problem as well. So, it can be another approach for the question.
Calculation is an important side of this problem. Don’t confuse with the calculation part of the problem as well.

Bookmark added to your notes.
View Notes