Question
Question: Find the second derivative of the following function. \[y=\dfrac{x\ln x}{1-{{x}^{2}}}\]...
Find the second derivative of the following function. y=1−x2xlnx
Solution
- Hint: Use product and quotient rule to differentiate the given function twice. Also use the fact that differentiation of y=lnx is dxdy=x1 and differentiation of y=axn+b is dxdy=anxn−1.
Complete step-by-step solution -
To find the second derivative of the function y=1−x2xlnx, we will differentiate it twice with respect to the variable x.
To find the first derivative, we will differentiate the function y with respect to x.
We will use the product rule of differentiation of product of two functions which states that if w(x)=u(x)v(x) then, we have dxdw(x)=dxd(u(x)×v(x))=v(x)×dxdu(x)+u(x)dxdv(x).
Substituting u(x)=1−x2x,v(x)=lnx in the above equation, we have dxdw(x)=v(x)×dxdu(x)+u(x)dxdv(x)=lnx×dxd(1−x2x)+1−x2x×dxd(lnx).....(1).
We know that differentiation of function y=lnx is dxdy=x1.....(2).
To find the value of dxd(1−x2x), we have u(x)=1−x2x.
We will use quotient rule of differentiation which states that if y=β(x)α(x) then, we have dxdy=β2(x)β(x)α′(x)−α(x)β′(x).
Substituting α(x)=x,β(x)=1−x2 in the above equation, we have dxdu(x)=β2(x)β(x)α′(x)−α(x)β′(x)=(1−x2)2(1−x2)×dxd(x)−x×dxd(1−x2).....(3).
We know that differentiation of any function of the form y=axn+b is dxdy=anxn−1.
Substituting a=1,n=1,b=0 in the above equation, we have dxd(x)=1.....(4).
Substituting a=−1,n=2,b=1 in the above equation, we have dxd(1−x2)=−2x.....(5).
Substituting equation (4) and (5) in equation (3), we have dxdu(x)=(1−x2)2(1−x2)×dxd(x)−x×dxd(1−x2)=(1−x2)2(1−x2)×1−x×(−2x)=(1−x2)21+x2.....(6).
Substituting equation (2) and (6) in equation (1), we get dxdw(x)=lnx×dxd(1−x2x)+1−x2x×dxd(lnx)=lnx×(1−x2)21+x2+1−x2x×x1=(1−x2)2lnx(1+x2)+1−x21.
Solving the above equation by taking LCM, we get dxdw(x)=(1−x2)2lnx(1+x2)+1−x21=(1−x2)2lnx(1+x2)+1−x2.
Thus, we have dxdy=(1−x2)2lnx(1+x2)+1−x2 as the first derivative of the given function.
Now, to find the second derivative, we will again differentiate the function dxdy as dx2d2y=dxd(dxdy).
Let’s assume a(x)=dxdy=(1−x2)2lnx(1+x2)+1−x2.
So, we have dx2d2y=dxda(x).
We can rewrite a(x) as a quotient of two functions d(x)c(x) where c(x)=lnx(1+x2)+1−x2,d(x)=(1−x2)2.
We will use quotient rule of differentiation which states that if y=d(x)c(x) then, we have dxdy=d2(x)d(x)c′(x)−c(x)d′(x).
Substituting c(x)=lnx(1+x2)+1−x2,d(x)=(1−x2)2 in the above equation, we get dxda(x)=d2(x)d(x)c′(x)−c(x)d′(x)=(1−x2)4(1−x2)2×dxd(lnx(1+x2)+1−x2)−(lnx(1+x2)+1−x2)×dxdy(1−x2)2.....(7).
To find the value of dxdy(1−x2)2, let’s assume d(x)=(1−x2)2 as a composition of two functions γ(β(x)) where γ(x)=x2,β(x)=1−x2.
We will use chain rule of differentiation of composition of two functions to find the derivative which states that if y is a composition of two functions γ(x) and β(x) such that y=γ(β(x))), then dxdy=dβ(x)dγ(β(x))×dxdβ(x).
Substituting γ(x)=x2,β(x)=1−x2 in the above equation, we get dxdd(x)=dβ(x)dγ(β(x))×dxdβ(x)=d(1−x2)d(1−x2)2×dxd(1−x2).....(8).
To find the value of d(1−x2)d(1−x2)2, let’s assume t=1−x2.
Thus, we have d(1−x2)d(1−x2)2=dtd(t2).
We know that differentiation of any function of the form y=axn+b is dxdy=anxn−1.
Substituting a=1,n=2,b=0 in the above equation, we get dtd(t2)=2t.
So, we have d(1−x2)d(1−x2)2=dtd(t2)=2t=2(1−x2).....(9).
Substituting equation (5) and (9) in equation (8), we get dxdd(x)=d(1−x2)d(1−x2)2×dxd(1−x2)=2(1−x2)×(−2x)=−4x(1−x2).....(10).
To find the value of dxd(lnx(1+x2)+1−x2), we can write c(x)=lnx(1+x2)+1−x2 as a sum of two functions c(x)=f(x)+β(x) where f(x)=lnx(1+x2),β(x)=1−x2.
We will use sum rule of differentiation of two functions which states that if y=f(x)+β(x) then dxdy=dxdf(x)+dxdβ(x).
Substituting f(x)=lnx(1+x2),β(x)=1−x2 in the above equation, we have dxdc(x)=dxdf(x)+dxdβ(x)=dxd(lnx(1+x2))+dxd(1−x2).....(11).
To find the value of dxd(lnx(1+x2)), we can write f(x)=lnx(1+x2) as a product of two functions f(x)=g(x)×h(x) where g(x)=lnx,h(x)=1+x2.
We will use the product rule of differentiation of product of two functions which states that if f(x)=g(x)h(x) then, we have dxdf(x)=dxd(g(x)×h(x))=g(x)×dxdh(x)+h(x)dxdg(x).
Substituting g(x)=lnx,h(x)=1+x2 in the above equation, we have dxdf(x)=g(x)×dxdh(x)+h(x)dxdg(x)=lnx×dxd(1+x2)+(1+x2)×dxd(lnx).....(12).
To find the value of dxd(1+x2), we will substitute a=1,n=2,b=1 in the formula where differentiation of y=axn+b is dxdy=anxn−1.
Thus, we have dxd(1+x2)=2x.....(13).
Substituting equation (2) and (13) in equation (12), we get dxdf(x)=lnx×dxd(1+x2)+(1+x2)×dxd(lnx)=lnx×(2x)+(1+x2)×x1.
Solving the above equation, we get dxdf(x)=lnx×(2x)+(1+x2)×x1=2xlnx+x1+x.....(14).
Substituting equation (5) and (14) in equation (11), we get dxdc(x)=dxd(lnx(1+x2))+dxd(1−x2)=2xlnx+x1+x−2x=2xlnx+x1−x.....(15).
Substituting equation (10) and (15) in equation (7), we get dxda(x)=(1−x2)4(1−x2)2×dxd(lnx(1+x2)+1−x2)−(lnx(1+x2)+1−x2)×dxdy(1−x2)2=(1−x2)4(1−x2)2(2xlnx+x1−x)−(lnx(1+x2)+1−x2)(−4x(1−x2)).
Solving the above equation, we get dxda(x)=(1−x2)4(1−x2)2(2xlnx+x1−x)−(lnx(1+x2)+1−x2)(−4x(1−x2))=(1−x2)22xlnx+x1−x+(1−x2)34x(lnx(1+x2)+1−x2).
Thus, we have dx2d2y=dxda(x)=(1−x2)22xlnx+x1−x+(1−x2)34x(lnx(1+x2)+1−x2).
Hence, the second derivative of y=1−x2xlnx is (1−x2)22xlnx+x1−x+(1−x2)34x(lnx(1+x2)+1−x2).
Note: It is necessary to use the chain rule of differentiation for composition of two functions. Otherwise, we won’t be able to find the second derivative of the given function. Also, one must know that the first derivative of any function is the slope of the function.