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

Find the derivative of the following y=x1−x1+x2y=x\sqrt{\dfrac{1-x}{1+{{x}^{2}}}}.

Explanation

Solution

Hint: To solve this type of question we have to use these two following formulas:
d(u.v)dx=uv′+vu′\dfrac{d(u.v)}{dx}=uv'+vu' and ddx(uv)=v.u′−u.v′v2\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)y=f(x).g(x) where f(x)=xf(x)=x and g(x)=1−x1+x2g(x)=\sqrt{\dfrac{1-x}{1+{{x}^{2}}}}.
To find the derivative of yy , we will apply the product rule of differentiation.
Now , we know , the product rule of differentiation is given as
y′=ddx(f(x).g(x))=g(x).f′(x)+f(x).g′(x)y'=\dfrac{d}{dx}(f(x).g(x))=g(x).{{f}^{'}}(x)+f(x).{{g}^{'}}(x)
Now , to calculate the value of y′y' , first we need to find the value of f′(x){{f}^{'}}(x)and g′(x){{g}^{'}}(x).
We have f(x)=xf(x)=x
So , on differentiating f(x)f(x) with respect to xx , we get ,
f′(x)=ddx(x)=1{{f}^{'}}(x)=\dfrac{d}{dx}(x)=1.
Again , we have g(x)=1−x1+x2g(x)=\sqrt{\dfrac{1-x}{1+{{x}^{2}}}}.
Now , let 1−x1+x2=p(x)\dfrac{1-x}{1+{{x}^{2}}}=p(x).
So , g(x)=p(x)g(x)=\sqrt{p(x)}
Now , ∵g(x)\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)h(x) is a composite function given by h(x)=f(g(x))h(x)=f(g(x)) , thenh′(x)=f′(g(x))×g′(x)h'(x)=f'(g(x))\times g'(x).”
So , g′(x)=12p(x).p′(x){{g}^{'}}(x)=\dfrac{1}{2\sqrt{p(x)}}.{{p}^{'}}(x).
Now , we will find p′(x){{p}^{'}}(x).
For this we will use the quotient rule , which is given as
ddx(uv)=v.u′−u.v′v2\dfrac{d}{dx}(\dfrac{u}{v})=\dfrac{v.{{u}^{'}}-u.v'}{{{v}^{2}}}
In p(x)p(x) , we have u=(1−x)u=(1-x) and v=(1+x2)v=(1+{x^2}) .
So , we get p′(x)=[−1×(1+x2)]−[(1−x)×2x](1+x2)2p'(x)=\dfrac{[-1\times (1+{{x}^{2}})]-[(1-x)\times 2x]}{{{(1+{{x}^{2}})}^{2}}}
=−1−x2−2x+2x2(1+x2)2=\dfrac{-1-{{x}^{2}}-2x+2{{x}^{2}}}{{{(1+{{x}^{2}})}^{2}}}
=x2−2x−1(1+x2)2=\dfrac{{{x}^{2}}-2x-1}{{{(1+{{x}^{2}})}^{2}}}
Now , we will substitute the value of p′(x){{p}^{'}}(x) in g′(x){{g}^{'}}(x).
On substituting the value of p′(x){{p}^{'}}(x) in g′(x){{g}^{'}}(x) we get,
g′(x)=121−x1+x2.x2−2x−1(1+x2)2{{g}^{'}}(x)=\dfrac{1}{2\sqrt{\dfrac{1-x}{1+{{x}^{2}}}}}.\dfrac{{{x}^{2}}-2x-1}{{{(1+{{x}^{2}})}^{2}}}
=1+x221−x.x2−2x−1(1+x2)2=\dfrac{\sqrt{1+{{x}^{2}}}}{2\sqrt{1-x}}.\dfrac{{{x}^{2}}-2x-1}{{{(1+{{x}^{2}})}^{2}}}
=x2−2x−12(1−x)12(1+x2)32=\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){{g}^{'}}(x) and f′(x){{f}^{'}}(x) in y′{{y}^{'}}.
On substituting the values of g′(x){{g}^{'}}(x) and f′(x){{f}^{'}}(x) in y′{{y}^{'}} , we get ,
y′=[1×1−x1+x2]+x×x2−2x−1(2)(1−x)12(1+x2)32{{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′=2(1−x)121−x.(1+x2)2(1−x)121+x2.(1+x2)+x(x2−2x−1)2(1−x)12(1+x2)32y'=\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}}}}
=2(1−x)(1+x2)+(x3−2x2−x)2(1−x)(1+x2)3=\dfrac{2(1-x)(1+{{x}^{2}})+({{x}^{3}}-2{{x}^{2}}-x)}{2\sqrt{(1-x){{(1+{{x}^{2}})}^{3}}}}
=2(1+x2−x−x3)+(x3−2x2−x)2(1−x)(1+x2)32=\dfrac{2(1+{{x}^{2}}-x-{{x}^{3}})+({{x}^{3}}-2{{x}^{2}}-x)}{2\sqrt{(1-x)}{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}}
=2+2x2−2x−2x3+x3−2x2−x2(1−x).(1+x2)32=\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}}}}
=−x3−3x+22(1−x).(1+x2)32=\dfrac{-{{x}^{3}}-3x+2}{2\sqrt{(1-x)}.{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}}
Hence, the derivative of yy is given as y′=−x3−3x+22(1−x).(1+x2)32y'=\dfrac{-{{x}^{3}}-3x+2}{2\sqrt{(1-x)}.{{(1+{{x}^{2}})}^{\dfrac{3}{2}}}}.
Note: Since g(x)g(x) is a composite function , g′(x)=12p(x).p′(x){{g}^{'}}(x)=\dfrac{1}{2\sqrt{p(x)}}.{{p}^{'}}(x).
Most of the students generally consider g(x)g(x)as a normal function and write g′(x)=12p(x){{g}^{'}}(x)=\dfrac{1}{2\sqrt{p(x)}} , which is wrong. This will result in the wrong calculation of y′y'. Hence such mistakes should be avoided.