Question

Find the derivative of the following \[y=x\sqrt{\dfrac{1-x}{1+{{x}^{2}}}}\].

Answer 1

Hint: To solve this type of question we have to use these two following formulas:
\[\dfrac{d(u.v)}{dx}=uv'+vu'\] and \[\dfrac{d}{dx}(\dfrac{u}{v})=\dfrac{v.{{u}^{'}}-u.v'}{{{v}^{2}}}\].
Complete step-by-step answer:
The given function is of the form \[y=f(x).g(x)\] where \[f(x)=x\] and \[g(x)=\sqrt{\dfrac{1-x}{1+{{x}^{2}}}}\].
To find the derivative of \[y\] , we will apply the product rule of differentiation.
Now , we know , the product rule of differentiation is given as
\[y'=\dfrac{d}{dx}(f(x).g(x))=g(x).{{f}^{'}}(x)+f(x).{{g}^{'}}(x)\]
Now , to calculate the value of \[y'\] , first we need to find the value of \[{{f}^{'}}(x)\]and \[{{g}^{'}}(x)\].
We have \[f(x)=x\]
So , on differentiating \[f(x)\] with respect to \[x\] , we get ,
\[{{f}^{'}}(x)=\dfrac{d}{dx}(x)=1\].
Again , we have \[g(x)=\sqrt{\dfrac{1-x}{1+{{x}^{2}}}}\].
Now , let \[\dfrac{1-x}{1+{{x}^{2}}}=p(x)\].
So , \[g(x)=\sqrt{p(x)}\]
Now , \[\because g(x)\] is a composite function , we will apply chain rule of differentiation. The chain rule of differentiation is given as: “If \[h(x)\] is a composite function given by \[h(x)=f(g(x))\] , then\[h'(x)=f'(g(x))\times g'(x)\].”
So , \[{{g}^{'}}(x)=\dfrac{1}{2\sqrt{p(x)}}.{{p}^{'}}(x)\].
Now , we will find \[{{p}^{'}}(x)\].
For this we will use the quotient rule , which is given as
\[\dfrac{d}{dx}(\dfrac{u}{v})=\dfrac{v.{{u}^{'}}-u.v'}{{{v}^{2}}}\]
In \[p(x)\] , we have \[u=(1-x)\] and \[v=(1+{x^2})\] .
So , we get \[p'(x)=\dfrac{[-1\times (1+{{x}^{2}})]-[(1-x)\times 2x]}{{{(1+{{x}^{2}})}^{2}}}\]
\[=\dfrac{-1-{{x}^{2}}-2x+2{{x}^{2}}}{{{(1+{{x}^{2}})}^{2}}}\]
\[=\dfrac{{{x}^{2}}-2x-1}{{{(1+{{x}^{2}})}^{2}}}\]
Now , we will substitute the value of \[{{p}^{'}}(x)\] in \[{{g}^{'}}(x)\].
On substituting the value of \[{{p}^{'}}(x)\] in \[{{g}^{'}}(x)\] we get,
\[{{g}^{'}}(x)=\dfrac{1}{2\sqrt{\dfrac{1-x}{1+{{x}^{2}}}}}.\dfrac{{{x}^{2}}-2x-1}{{{(1+{{x}^{2}})}^{2}}}\]
\[=\dfrac{\sqrt{1+{{x}^{2}}}}{2\sqrt{1-x}}.\dfrac{{{x}^{2}}-2x-1}{{{(1+{{x}^{2}})}^{2}}}\]
\[=\dfrac{{{x}^{2}}-2x-1}{2{{(1-x)}^{\dfrac{1}{2}}}{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}}\]
Now , we will substitute the values of \[{{g}^{'}}(x)\] and \[{{f}^{'}}(x)\] in \[{{y}^{'}}\].
On substituting the values of \[{{g}^{'}}(x)\] and \[{{f}^{'}}(x)\] in \[{{y}^{'}}\] , we get ,
\[{{y}^{'}}=[1\times \sqrt{\dfrac{1-x}{1+{{x}^{2}}}}]+x\times \dfrac{{{x}^{2}}-2x-1}{(2){{(1-x)}^{\dfrac{1}{2}}}{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}}\]
Now , we will take the LCM of the denominators.
On taking LCM of the denominators , we get ,
\[y'=\dfrac{2{{(1-x)}^{\dfrac{1}{2}}}\sqrt{1-x}.(1+{{x}^{2}})}{2{{(1-x)}^{\dfrac{1}{2}}}\sqrt{1+{{x}^{2}}}.(1+{{x}^{2}})}+\dfrac{x({{x}^{2}}-2x-1)}{2{{(1-x)}^{\dfrac{1}{2}}}{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}}\]
\[=\dfrac{2(1-x)(1+{{x}^{2}})+({{x}^{3}}-2{{x}^{2}}-x)}{2\sqrt{(1-x){{(1+{{x}^{2}})}^{3}}}}\]
\[=\dfrac{2(1+{{x}^{2}}-x-{{x}^{3}})+({{x}^{3}}-2{{x}^{2}}-x)}{2\sqrt{(1-x)}{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}}\]
\[=\dfrac{2+2{{x}^{2}}-2x-2{{x}^{3}}+{{x}^{3}}-2{{x}^{2}}-x}{2\sqrt{(1-x)}.{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}}\]
\[=\dfrac{-{{x}^{3}}-3x+2}{2\sqrt{(1-x)}.{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}}\]
Hence, the derivative of \[y\] is given as \[y'=\dfrac{-{{x}^{3}}-3x+2}{2\sqrt{(1-x)}.{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}}\].
Note: Since \[g(x)\] is a composite function , \[{{g}^{'}}(x)=\dfrac{1}{2\sqrt{p(x)}}.{{p}^{'}}(x)\].
Most of the students generally consider \[g(x)\]as a normal function and write \[{{g}^{'}}(x)=\dfrac{1}{2\sqrt{p(x)}}\] , which is wrong. This will result in the wrong calculation of \[y'\]. Hence such mistakes should be avoided.