Question

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

Answer 1

Hint: 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: \[{{\left[ v{{\left( x \right)}^{n}} \right]}^{\prime }}=nv{{\left( x \right)}^{n-1}}.{v}'\left( x \right)\]