Question

How to find the derivative of $x\sqrt{1-x}$?

Answer 1

Apply the product rule in the given function $x\sqrt{1-x}$and the formula of product rule is: ${{\left[ u\left( x \right).v\left( x \right) \right]}^{\prime }}=u\left( x \right).{v}'\left( x \right)+v\left( x \right).{u}'\left( x \right)$ where $u(x)=x$and $v(x)=\sqrt{1-x}$. To further differentiate $x$apply differentiation rule that is ${u}'\left( x \right)=n{{x}^{n-1}}$ where according to the question $n$ is equal to $1$and to differentiate $\sqrt{1-x}$ apply power rule: ${{\left[ v{{\left( x \right)}^{n}} \right]}^{\prime }}=nv{{\left( x \right)}^{n-1}}.{v}'\left( x \right)$ where according to the question $n$ is equal to $\dfrac{1}{2}$.

Complete step by step solution:
The derivative of $x\sqrt{1-x}$ is as follows:
$\dfrac{d}{dx}\left[ x\sqrt{1-x} \right]$
Applying product rule: ${{\left[ u\left( x \right).v\left( x \right) \right]}^{\prime }}=u\left( x \right).{v}'\left( x \right)+v\left( x \right).{u}'\left( x \right)$ in the given function we get:
$\Rightarrow \dfrac{d}{dx}\left[ x \right]\sqrt{1-x}+x\dfrac{d}{dx}\left[ \sqrt{1-x} \right]...(i)$
Now to further differentiate $x$apply differentiation rule that is ${u}'\left( x \right)=n{{x}^{n-1}}$
$\Rightarrow \dfrac{d}{dx}\left[ x \right]=1...(ii)$
And to differentiate $\sqrt{1-x}$ apply power rule: ${{\left[ v{{\left( x \right)}^{n}} \right]}^{\prime }}=nv{{\left( x \right)}^{n-1}}.{v}'\left( x \right)$ $\Rightarrow \dfrac{d}{dx}\left[ \sqrt{1-x} \right]=\dfrac{1}{2}{{\left( 1-x \right)}^{\dfrac{1}{2}-1}}\cdot \dfrac{d}{dx}\left[ 1-x \right]...(iii)$
Now putting the value of equation $(ii)$and $(iii)$ in equation $(i)$ we get:
$\Rightarrow 1\times \sqrt{1-x}+x\times \dfrac{1}{2}{{\left( 1-x \right)}^{\dfrac{1}{2}-1}}\times \dfrac{d}{dx}\left[ 1-x \right]$
Taking the LCM of the power of $\left( 1-x \right)$ that is $\dfrac{1}{2}-1=\dfrac{1-2}{2}=-\dfrac{1}{2}$ we get power of $\left( 1-x \right)$ equal to $-\dfrac{1}{2}$ and putting this in the above equation we get:
$\Rightarrow \sqrt{1-x}+\dfrac{x}{2}{{\left( 1-x \right)}^{-\dfrac{1}{2}}}\times \dfrac{d}{dx}\left[ 1-x \right]$
We can also write ${{\left( 1-x \right)}^{-\dfrac{1}{2}}}$ as $\dfrac{1}{\sqrt{1-x}}$. Applying it to the above equation we get:
$\Rightarrow \sqrt{1-x}+\dfrac{x}{2\sqrt{1-x}}\times \dfrac{d}{dx}\left[ 1-x \right]$
After solving the terms within the brackets, we get:
$\Rightarrow \sqrt{1-x}+\dfrac{x\left( \dfrac{d}{dx}\left[ 1 \right]-\dfrac{d}{dx}\left[ x \right] \right)}{2\sqrt{1-x}}$
We know that derivative of $1$ is $0$ and for the derivative of $x$ again apply differentiation rule that is ${u}'\left( x \right)=n{{x}^{n-1}}$ that is $\dfrac{d}{dx}\left[ x \right]=1$
$\Rightarrow \sqrt{1-x}+\dfrac{x\left( 0-1 \right)}{2\sqrt{1-x}}$
After solving the terms within the brackets, we get:
$\Rightarrow \sqrt{1-x}-\dfrac{x}{2\sqrt{1-x}}$
By simplifying it further & taking the LCM we get:
$\Rightarrow \dfrac{2(1-x)-x}{2\sqrt{1-x}}$
After solving the brackets in the numerator, we get:
$\Rightarrow -\dfrac{3x-2}{2\sqrt{1-x}}$
$\therefore$Derivative of $x\sqrt{1-x}$ is $-\dfrac{3x-2}{2\sqrt{1-x}}$.

Note:
Students can go wrong by not applying power rule in the function $\sqrt{1-x}$ correctly that is they write $\dfrac{d}{dx}\left[ \sqrt{1-x} \right]=\dfrac{1}{2}{{\left( \sqrt{1-x} \right)}^{\dfrac{1}{2}-1}}$and forget to multiply with the derivative of $\left( 1-x \right)$ which further leads to the wrong answer. So, the key point is to know all differentiation rule, power rule & product rule right that is the differentiation rule: ${u}'\left( x \right)=n{{x}^{n-1}}$, the product rule: ${{\left[ u\left( x \right).v\left( x \right) \right]}^{\prime }}=u\left( x \right).{v}'\left( x \right)+v\left( x \right).{u}'\left( x \right)$ & power rule: ${{\left[ v{{\left( x \right)}^{n}} \right]}^{\prime }}=nv{{\left( x \right)}^{n-1}}.{v}'\left( x \right)$