Second Order Derivative

We have already learned from the previous lecture on First Order Derivative Solution that if any function y=f(x) be derivable at any interval with respect to x then dy/dx=f’(x) represents first order differential of y with respect to x. Again if f’(x) also becomes derivable in that interval with respect to x then we can write,
\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\left( \frac{dy}{dx} \right)=\frac{d}{dx}\left( f'(x) \right)=f”(x)\]
This represents second order differential of y with respect to x.

2^nd Order Derivative in terms of Limit

We already learned if y=f(x) then,
\[\frac{dy}{dx}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(x+h)-f(x)}{h}=f'(x)\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\left( \frac{dy}{dx} \right)=\frac{d}{dx}\left( f'(x) \right)\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=\underset{h\to 0}{\mathop{\lim }}\,\frac{f'(x+h)-f'(x)}{h}\]
\[=\underset{h\to 0}{\mathop{\lim }}\,\frac{\frac{f(x+2h)-f(x+h)}{h}-\frac{f(x+h)-f(x)}{h}}{h}\]
\[=\underset{h\to 0}{\mathop{\lim }}\,\frac{f(x+2h)-2f(x+h)+f(x)}{{{h}^{2}}}\]

Geometrical Significance of 2^nd Order Derivative

From our First order derivative Solution lecture we have learned that f’(x) represents ‘Rate of change of y with respect to x’. Again f’(x) is ‘Slope of the tangent’ of the curve y=f(x).

Hence we can justify f’’(x) as the ‘Rate of change of slope of the tangent’ of the curve y=f(x).

For more info go to Wikipedia

\[If,x=a(t-\sin t),y=a(1+\cos t)then\]
\[find\frac{{{d}^{2}}y}{d{{x}^{2}}}at-t=\frac{\pi }{2}\]
Solution:
\[x=a(t-\sin t)\]
\[\therefore \frac{dx}{dt}=a(1-\cos t)\]
\[y=a(1+\cos t)\]
\[\therefore \frac{dy}{dt}=-a\sin t\]
\[\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{-a\sin t}{a(1-\cos t)}=-\frac{a.2\sin \frac{t}{2}\cos \frac{t}{2}}{a.2{{\sin }^{2}}\frac{t}{2}}=-\cot \frac{t}{2}\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=-\frac{d}{dx}\left( \cot \frac{t}{2} \right)=-\frac{d}{dy}\left( \cot \frac{t}{2} \right).\frac{dt}{dx}\]
\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=-\left( -\cos e{{c}^{2}}\frac{t}{2} \right).\frac{1}{2}.\frac{1}{a(1-\cos t)}\]
\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{\cos e{{c}^{2}}\frac{t}{2}}{2.a.2{{\sin }^{2}}\frac{t}{2}}=\frac{1}{4a}.\cos e{{c}^{4}}\frac{t}{2}\]
\[\therefore {{\left[ \frac{{{d}^{2}}y}{d{{x}^{2}}} \right]}_{t=\frac{\pi }{2}}}=\frac{1}{4a}.{{\left( \cos ec\frac{\pi }{4} \right)}^{4}}=\frac{1}{4a}{{\left( \sqrt{2} \right)}^{4}}=\frac{1}{a}\]

\[If,y=\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}},show-that\]
\[\left( 1-{{x}^{2}} \right)\frac{{{d}^{2}}y}{d{{x}^{2}}}-3x\frac{dy}{dx}-y=0\]
Solution:
\[y=\frac{{{\sin }^{-1}}x}{\sqrt{1-{{x}^{2}}}}\Rightarrow \sqrt{1-{{x}^{2}}}.y={{\sin }^{-1}}x\]
Squaring both sides we get,
\[\left( 1-{{x}^{2}} \right){{y}^{2}}={{\left( {{\sin }^{-1}}x \right)}^{2}}\]
\[\therefore \left( 1-{{x}^{2}} \right).2y\frac{dy}{dx}+(-2x){{y}^{2}}=2.{{\sin }^{-1}}x.\frac{1}{\sqrt{1-{{x}^{2}}}}\]
\[\Rightarrow 2\left( 1-{{x}^{2}} \right)y\frac{dy}{dx}-2x{{y}^{2}}=2.y\]
\[\Rightarrow \left( 1-{{x}^{2}} \right)\frac{dy}{dx}-xy=1\]
\[\therefore \left( 1-{{x}^{2}} \right)\frac{{{d}^{2}}y}{d{{x}^{2}}}-2x\frac{dy}{dx}-\left( x\frac{dy}{dx}+1.y \right)=0\]
\[\therefore \left( 1-{{x}^{2}} \right)\frac{{{d}^{2}}y}{d{{x}^{2}}}-3x\frac{dy}{dx}-y=0\]

\[If,y=\sin \left( 2{{\sin }^{-1}}x) \right)show-that\]
\[(1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}+4y=0\]
Solution:
\[y=\sin \left( 2{{\sin }^{-1}}x) \right)…………..(1)\]
\[\therefore \frac{dy}{dx}=\cos \left( 2{{\sin }^{-1}}x) \right).2.\frac{1}{\sqrt{1-{{x}^{2}}}}\]
\[\Rightarrow \sqrt{1-{{x}^{2}}}\frac{dy}{dx}=2\cos \left( 2{{\sin }^{-1}}x) \right)\]
Squaring both sides we get,
\[(1-{{x}^{2}}){{\left( \frac{dy}{dx} \right)}^{2}}=4{{\cos }^{2}}\left( 2{{\sin }^{-1}}x) \right)\]
\[\Rightarrow (1-{{x}^{2}}){{\left( \frac{dy}{dx} \right)}^{2}}=4\left[ 1-{{\sin }^{2}}\left( 2{{\sin }^{-1}}x) \right) \right]\]
\[\Rightarrow (1-{{x}^{2}}){{\left( \frac{dy}{dx} \right)}^{2}}=4(1-{{y}^{2}})\]
Differentiating both sides w.r.t. x we get,
\[(1-{{x}^{2}}).2\frac{dy}{dx}.\frac{{{d}^{2}}y}{d{{x}^{2}}}+(-2x).{{\left( \frac{dy}{dx} \right)}^{2}}=4.\left( -2y\frac{dy}{dx} \right)\]
\[\Rightarrow (1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}=-4y,\left[ \because \frac{dy}{dx}\ne 0 \right]\]
\[\therefore (1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}+4y=0\]

\[If,y=\sin \left( m{{\sin }^{-1}}x) \right)show-that\]
\[(1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}+{{m}^{2}}y=0\]
Solution:
Do the same as above example.

\[If,y={{e}^{m{{\sin }^{-1}}x}}show-that\]
\[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}={{m}^{2}}y\]
Solution:
\[y={{e}^{m{{\sin }^{-1}}x}}\]
\[\therefore \frac{dy}{dx}={{e}^{m{{\sin }^{-1}}x}}.m.\frac{1}{\sqrt{1-{{x}^{2}}}}=\frac{my}{\sqrt{1-{{x}^{2}}}}\]
\[\Rightarrow \sqrt{1-{{x}^{2}}}\frac{dy}{dx}=my\]
\[\Rightarrow (1-{{x}^{2}}){{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}{{y}^{2}}\]
\[\therefore (1-{{x}^{2}}).2\frac{dy}{dx}.\frac{{{d}^{2}}y}{d{{x}^{2}}}+(-2x){{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}.2y\frac{dy}{dx}\]
\[\Rightarrow (1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}={{m}^{2}}y\]

\[If,y=a\cos ({{\log }_{e}}x)+b\sin ({{\log }_{e}}x)show\]
\[that,{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}+y=0\]
\[y=a\cos ({{\log }_{e}}x)+b\sin ({{\log }_{e}}x)\]
\[\therefore \frac{dy}{dx}=-a\sin ({{\log }_{e}}x).\frac{1}{x}+b\cos ({{\log }_{e}}x).\frac{1}{x}\]
\[\Rightarrow x\frac{dy}{dx}=-a\sin ({{\log }_{e}}x)+b\cos ({{\log }_{e}}x)\]
\[\therefore x\frac{{{d}^{2}}y}{d{{x}^{2}}}+1.\frac{dy}{dx}=-a\cos ({{\log }_{e}}x).\frac{1}{x}-b\sin ({{\log }_{e}}x).\frac{1}{x}\]
\[\Rightarrow {{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}=-\left[ a\cos ({{\log }_{e}}x)+b\sin ({{\log }_{e}}x) \right]\]
\[\Rightarrow {{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}=-y\]
\[\therefore {{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}+y=0\]

\[If,F(x)=f(x)\phi (x),f'(x)\phi ‘(x)=a,show\]
\[that,\frac{F(x)}{F'(x)}=\frac{f”(x)}{f(x)}+\frac{\phi ”(x)}{\phi (x)}+\frac{2a}{f(x)\phi (x)}\]
\[i.e.,\frac{F}{F’}=\frac{f”}{f}+\frac{\phi ”}{\phi }+\frac{2a}{f\phi }\]
Solution:
\[F(x)=f(x)\phi (x)\]
\[\therefore F'(x)=f(x)\phi ‘(x)+f'(x)\phi (x)\]
\[\therefore F”(x)=f(x)\phi ”(x)+f'(x)\phi ‘(x)+f'(x)\phi ‘(x)+f”(x)\phi (x)\]
\[\Rightarrow F”(x)=f”(x)\phi (x)+f(x)\phi ”(x)+2f'(x)\phi ‘(x)\]
\[\therefore \frac{F”(x)}{F(x)}=\frac{f”(x)\phi (x)+f(x)\phi ”(x)+2f'(x)\phi ‘(x)}{F(x)}\]
\[\Rightarrow \frac{F”(x)}{F(x)}=\frac{f”(x)\phi (x)+f(x)\phi ”(x)+2a}{f(x)\phi (x)}\]
\[\Rightarrow \frac{F”(x)}{F(x)}=\frac{f”(x)\phi (x)}{f(x)\phi (x)}+\frac{f(x)\phi ”(x)}{f(x)\phi (x)}+\frac{2a}{f(x)\phi (x)}\]
\[\therefore \frac{F(x)}{F'(x)}=\frac{f”(x)}{f(x)}+\frac{\phi ”(x)}{\phi (x)}+\frac{2a}{f(x)\phi (x)}\]
\[i.e.,\frac{F}{F’}=\frac{f”}{f}+\frac{\phi ”}{\phi }+\frac{2a}{f\phi }\]

\[If,y=f(x),x=\frac{1}{z},show-that\]
\[\frac{{{d}^{2}}f}{d{{x}^{2}}}=2{{z}^{2}}\frac{dy}{dz}+{{z}^{4}}\frac{{{d}^{2}}y}{d{{z}^{2}}}\]
Solution:
\[y=f(x),x=\frac{1}{z}\]
\[\therefore \frac{dx}{dz}=-\frac{1}{{{z}^{2}}}\Rightarrow \frac{dz}{dx}=-{{z}^{2}}……….(1)\]
\[\frac{dy}{dx}=\frac{dy}{dz}.\frac{dz}{dx}=-{{z}^{2}}\frac{dy}{dz}[by(1)]\]
\[\therefore \frac{d}{dx}\left( \frac{dy}{dx} \right)=-\frac{d}{dx}\left( {{z}^{2}}\frac{dy}{dz} \right)\]
\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=-\frac{d}{dz}\left( {{z}^{2}}\frac{dy}{dz} \right).\frac{dz}{dx}\]
\[\therefore \frac{{{d}^{2}}f}{d{{x}^{2}}}=2{{z}^{2}}\frac{dy}{dz}+{{z}^{4}}\frac{{{d}^{2}}y}{d{{z}^{2}}}\]
\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=-\left[ 2z.\frac{dy}{dz}+{{z}^{2}}\frac{{{d}^{2}}y}{d{{z}^{2}}} \right].\left( -{{z}^{2}} \right)\]
\[\therefore \frac{{{d}^{2}}f}{d{{x}^{2}}}=2{{z}^{2}}\frac{dy}{dz}+{{z}^{4}}\frac{{{d}^{2}}y}{d{{z}^{2}}}\]

\[If,x=\cos \theta ,y=\cos m\theta ,show-that\]
\[(1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}+{{m}^{2}}y=0\]
Solution:
\[x=\cos \theta ,y=\cos m\theta \]
\[\therefore \theta ={{\cos }^{-1}}x,\therefore y=\cos m\left( {{\cos }^{-1}}x \right)………(1)\]
\[\therefore \frac{dy}{dx}=-\sin m\left( {{\cos }^{-1}}x \right).m.\left( -\frac{1}{\sqrt{1-{{x}^{2}}}} \right)\]
\[\Rightarrow \sqrt{1-{{x}^{2}}}\frac{dy}{dx}=m\sin m\left( {{\cos }^{-1}}x \right)\]
Squaring both sides we get,
\[\left( 1-{{x}^{2}} \right){{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}{{\sin }^{2}}m\left( {{\cos }^{-1}}x \right)\]
\[\Rightarrow \left( 1-{{x}^{2}} \right){{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}\left[ 1-{{\cos }^{2}}m\left( {{\cos }^{-1}}x \right) \right]\]
\[\Rightarrow \left( 1-{{x}^{2}} \right){{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}\left[ 1-{{y}^{2}} \right],[by(1)]\]
\[\therefore \left( 1-{{x}^{2}} \right).2\frac{dy}{dx}.\frac{{{d}^{2}}y}{d{{x}^{2}}}-2x{{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}\left( -2y\frac{dy}{dx} \right)\]
\[\Rightarrow \left( 1-{{x}^{2}} \right)\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}=-{{m}^{2}}y\]
\[\therefore \left( 1-{{x}^{2}} \right)\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}+{{m}^{2}}y=0\]

\[If,x={{e}^{t}}\sin t,y={{e}^{t}}\cos t,show-that\]
\[{{(x+y)}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=2\left( x\frac{dy}{dx}-y \right)\]
Solution
\[x={{e}^{t}}\sin t,y={{e}^{t}}\cos t\]
\[\therefore \frac{dx}{dt}={{e}^{t}}\sin t+{{e}^{t}}\cos t;\frac{dy}{dt}={{e}^{t}}\cos t-{{e}^{t}}\sin t\]
\[\Rightarrow \frac{dx}{dt}=x+y;\frac{dy}{dt}=y-x\]
\[\therefore \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{y-x}{x+y}\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{(x+y)\left( \frac{dy}{dx}-1 \right)-(y-x)\left( 1+\frac{dy}{dx} \right)}{{{(x+y)}^{2}}}\]
\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{(x+y-y+x)\frac{dy}{dx}-(x+y+y-x)}{{{(x+y)}^{2}}}\]
\[\Rightarrow {{(x+y)}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=2x\frac{dy}{dx}-2y\]
\[\therefore {{(x+y)}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=2\left( x\frac{dy}{dx}-y \right)\]

\[If,2x={{y}^{\frac{1}{m}}}+{{y}^{-\frac{1}{m}}},show-that\]
\[\left( {{x}^{2}}-1 \right){{y}_{2}}+x{{y}_{1}}-{{m}^{2}}y=0\]
Solution:
\[2x={{y}^{\frac{1}{m}}}+{{y}^{-\frac{1}{m}}}……….(1)\]
\[\therefore 2.1=\frac{1}{m}.{{y}^{\frac{1}{m}-1}}.{{y}_{1}}-\frac{1}{m}.{{y}^{-\frac{1}{m}-1}}.{{y}_{1}}\]
\[\Rightarrow 2m=\left( \frac{{{y}^{\frac{1}{m}}}}{y}-\frac{{{y}^{-\frac{1}{m}}}}{y} \right){{y}_{1}}\]
\[\Rightarrow 2my=\left( {{y}^{\frac{1}{m}}}-{{y}^{-\frac{1}{m}}} \right){{y}_{1}}\]
Squaring both sides we get,
\[4{{m}^{2}}{{y}^{2}}={{\left( {{y}^{\frac{1}{m}}}-{{y}^{-\frac{1}{m}}} \right)}^{2}}{{({{y}_{1}})}^{2}}=\left[ {{\left( {{y}^{\frac{1}{m}}}+{{y}^{-\frac{1}{m}}} \right)}^{2}}-4{{y}^{\frac{1}{m}}}.{{y}^{-\frac{1}{m}}} \right]{{({{y}_{1}})}^{2}}\]
\[\Rightarrow 4{{m}^{2}}{{y}^{2}}=\left[ 4{{x}^{2}}-4 \right]{{({{y}_{1}})}^{2}}\Rightarrow ({{x}^{2}}-1){{({{y}_{1}})}^{2}}={{m}^{2}}{{y}^{2}}\]
\[\therefore ({{x}^{2}}-1).2{{y}_{1}}.{{y}_{2}}+2x.{{({{y}_{1}})}^{2}}={{m}^{2}}.2y.{{y}_{1}}\]
\[\Rightarrow ({{x}^{2}}-1).{{y}_{2}}+x{{y}_{1}}={{m}^{2}}y\]
\[\therefore ({{x}^{2}}-1){{y}_{2}}+x{{y}_{1}}-{{m}^{2}}y=0\]

\[If,y=A{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{n}}+B{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{n}}show\]
\[that,({{x}^{2}}-1){{y}_{2}}+x{{y}_{1}}={{n}^{2}}y\]
Solution:
\[y=A{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{n}}+B{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{n}}\]
\[\therefore {{y}_{1}}=An{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{n-1}}.\left[ 1+\frac{2x}{2\sqrt{{{x}^{2}}-1}} \right]\]
\[+Bn{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{n-1}}.\left[ 1-\frac{2x}{2\sqrt{{{x}^{2}}-1}} \right]\]
\[\therefore {{y}_{1}}=An{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{n-1}}.\left[ \frac{x+\sqrt{{{x}^{2}}-1}}{\sqrt{{{x}^{2}}-1}} \right]\]
\[+Bn{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{n-1}}.\left[ \frac{\sqrt{{{x}^{2}}-1}-x}{\sqrt{{{x}^{2}}-1}} \right]\]
\[\Rightarrow {{y}_{1}}=\frac{An{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{n}}}{\sqrt{{{x}^{2}}-1}}-\frac{Bn{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{n}}}{\sqrt{{{x}^{2}}-1}}\]
\[\Rightarrow \sqrt{{{x}^{2}}-1}.{{y}_{1}}=An{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{n}}-Bn{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{n}}\]
Squaring both sides we get,
\[\left( {{x}^{2}}-1 \right){{({{y}_{1}})}^{2}}={{n}^{2}}{{\left[ A{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{n}}-B{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{n}} \right]}^{2}}\]
\[\Rightarrow \left( {{x}^{2}}-1 \right){{({{y}_{1}})}^{2}}={{n}^{2}}\left[ {{\left\{ A{{\left[ x+\sqrt{{{x}^{2}}-1} \right]}^{n}}+B{{\left[ x-\sqrt{{{x}^{2}}-1} \right]}^{n}} \right\}}^{2}}-4AB{{({{x}^{2}}-{{x}^{2}}+1)}^{n}} \right]\]
\[\left[ As,{{(a-b)}^{2}}={{(a+b)}^{2}}-4ab \right]\]
\[\Rightarrow \left( {{x}^{2}}-1 \right){{({{y}_{1}})}^{2}}={{n}^{2}}\left( {{y}^{2}}-4AB \right)\]
\[\therefore \left( {{x}^{2}}-1 \right).2{{y}_{1}}.{{y}_{2}}+2x{{({{y}_{1}})}^{2}}={{n}^{2}}.2y.{{y}_{1}}\]
\[\therefore ({{x}^{2}}-1){{y}_{2}}+x{{y}_{1}}={{n}^{2}}y\]

\[If,y=\cos \left[ m{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}+1} \right) \right]show\]
\[that,\left( {{x}^{2}}+1 \right)\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}+{{m}^{2}}y=0\]
Solution:
\[y=\cos \left[ m{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}+1} \right) \right]\]
\[\therefore \frac{dy}{dx}=-\sin \left[ m{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}+1} \right) \right].m.\frac{1+\frac{2x}{2\sqrt{{{x}^{2}}+1}}}{\left( x+\sqrt{{{x}^{2}}+1} \right)}\]
\[\therefore \frac{dy}{dx}=-m\sin \left[ m{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}+1} \right) \right].\frac{\frac{x+\sqrt{{{x}^{2}}+1}}{\sqrt{{{x}^{2}}+1}}}{\left( x+\sqrt{{{x}^{2}}+1} \right)}\]
\[\Rightarrow \frac{dy}{dx}=\frac{-m\sin \left[ m{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}+1} \right) \right]}{\sqrt{{{x}^{2}}+1}}\]
\[\Rightarrow \sqrt{{{x}^{2}}+1}\frac{dy}{dx}=-m\sin \left[ m{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}+1} \right) \right]\]
Squaring both sides we get,
\[\Rightarrow \left( {{x}^{2}}+1 \right){{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}{{\sin }^{2}}\left[ m{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}+1} \right) \right]\]
\[\Rightarrow \left( {{x}^{2}}+1 \right){{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}\left[ 1-{{\cos }^{2}}\left[ m{{\log }_{e}}\left( x+\sqrt{{{x}^{2}}+1} \right) \right] \right]\]
\[\Rightarrow \left( {{x}^{2}}+1 \right){{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}\left( 1-{{y}^{2}} \right)\]
\[\therefore \left( {{x}^{2}}+1 \right).2\frac{dy}{dx}.\frac{{{d}^{2}}y}{d{{x}^{2}}}+2x.{{\left( \frac{dy}{dx} \right)}^{2}}={{m}^{2}}\left( -2y\frac{dy}{dx} \right)\]
\[\Rightarrow \left( {{x}^{2}}+1 \right)\frac{{{d}^{2}}y}{d{{x}^{2}}}+x{{\frac{dy}{dx}}^{2}}=-{{m}^{2}}y\]
\[\therefore \left( {{x}^{2}}+1 \right)\frac{{{d}^{2}}y}{d{{x}^{2}}}+x\frac{dy}{dx}+{{m}^{2}}y=0\]

\[If,y={{\left( {{\sin }^{-1}}x \right)}^{2}}+{{\left( {{\cos }^{-1}}x \right)}^{2}}show\]
\[that,(1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}=4\]
Solution:
\[y={{\left( {{\sin }^{-1}}x \right)}^{2}}+{{\left( {{\cos }^{-1}}x \right)}^{2}}\]
\[\Rightarrow y={{\left( {{\sin }^{-1}}x+{{\cos }^{-1}}x \right)}^{2}}-2{{\sin }^{-1}}x.{{\cos }^{-1}}x\]
\[\Rightarrow y={{\left( \frac{\pi }{2} \right)}^{2}}-2{{\sin }^{-1}}x.\left( \frac{\pi }{2}-{{\sin }^{-1}}x \right)\]
\[\left[ \because {{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2} \right]\]
\[\Rightarrow y={{\left( \frac{\pi }{2} \right)}^{2}}-\pi {{\sin }^{-1}}x+2{{\left( {{\sin }^{-1}}x \right)}^{2}}\]
\[\therefore \frac{dy}{dx}=0-\frac{\pi }{\sqrt{1-{{x}^{2}}}}+2.2{{\sin }^{-1}}x.\frac{1}{\sqrt{1-{{x}^{2}}}}\]
\[\Rightarrow \sqrt{1-{{x}^{2}}}\frac{dy}{dx}=-\pi +4{{\sin }^{-1}}x\]
\[\therefore \sqrt{1-{{x}^{2}}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+\frac{-2x}{2\sqrt{1-{{x}^{2}}}}\frac{dy}{dx}=\frac{4}{\sqrt{1-{{x}^{2}}}}\]
\[\Rightarrow (1-{{x}^{2}})\frac{{{d}^{2}}y}{d{{x}^{2}}}-x\frac{dy}{dx}=4\]
\[\therefore (1-{{x}^{2}}){{y}_{2}}-x{{y}_{1}}=4\]

\[If,a{{x}^{2}}+2hxy+b{{y}^{2}}=1,show\]
\[that,\frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{{{h}^{2}}-ab}{{{\left( hx+by \right)}^{3}}}\]
\[a{{x}^{2}}+2hxy+b{{y}^{2}}=1……..(1)\]
\[\therefore 2ax+2h\left( 1.y+x\frac{dy}{dx} \right)+2by\frac{dy}{dx}=0\]
\[\Rightarrow \left( hx+by \right)\frac{dy}{dx}=-\left( ax+hy \right)\]
\[\Rightarrow \frac{dy}{dx}=-\frac{\left( ax+hy \right)}{\left( hx+by \right)}……..(2)\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=-\left[ \frac{\left( hx+by \right)\left( a+h\frac{dy}{dx} \right)-\left( ax+hy \right)\left( h+b\frac{dy}{dx} \right)}{{{\left( hx+by \right)}^{2}}} \right]\]

\[If,\cos x=y\cos (a+x),show-that\]
\[\frac{{{d}^{2}}y}{d{{x}^{2}}}=2\sin a{{\sec }^{2}}(a+x)\tan (a+x)\]
Solution:
\[\cos x=y\cos (a+x)\Rightarrow y=\frac{\cos x}{\cos (a+x)}\]
\[\therefore \frac{dy}{dx}=\frac{\cos (a+x).\left( -\sin a \right)-\cos x\left( -\sin (a+x) \right)}{{{\cos }^{2}}(a+x)}\]
\[\Rightarrow \frac{dy}{dx}=\frac{\sin (a+x)\cos x-\cos (a+x)\sin a}{{{\cos }^{2}}(a+x)}\]
\[\Rightarrow \frac{dy}{dx}=\frac{\sin (a+x-x)}{{{\cos }^{2}}(a+x)}=\sin a{{\sec }^{2}}(a+x)\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=\sin a.2\sec (a+x).\sec (a+x)\tan (a+x)\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=2\sin a{{\sec }^{2}}(a+x)\tan (a+x)\]

\[If,x=A{{e}^{-\frac{kt}{2}}}\cos (pt+\varepsilon ),show-that\]
\[\frac{{{d}^{2}}x}{d{{t}^{2}}}+k\frac{dx}{dt}+{{n}^{2}}x,where,{{n}^{2}}={{p}^{2}}+\frac{1}{4}{{k}^{2}}\]
Solution:
\[x=A{{e}^{-\frac{kt}{2}}}\cos (pt+\varepsilon )\]
\[\Rightarrow {{e}^{\frac{kt}{2}}}x=A\cos (pt+\varepsilon )……….(1)\]
\[\therefore {{e}^{\frac{kt}{2}}}.\frac{dx}{dt}+{{e}^{\frac{kt}{2}}}.\frac{k}{2}.x=-A\sin (pt+\varepsilon ).p\]
\[\Rightarrow {{e}^{\frac{kt}{2}}}.\frac{dx}{dt}+\frac{k}{2}{{e}^{\frac{kt}{2}}}x=-Ap\sin (pt+\varepsilon )\]
\[\therefore {{e}^{\frac{kt}{2}}}.\frac{{{d}^{2}}x}{d{{t}^{2}}}+{{e}^{\frac{kt}{2}}}.\frac{k}{2}.\frac{dx}{dt}+\frac{k}{2}\left( {{e}^{\frac{kt}{2}}}.\frac{dx}{dt}+\frac{k}{2}{{e}^{\frac{kt}{2}}}x \right)=-Ap\cos (pt+\varepsilon ).p\]
\[\Rightarrow {{e}^{\frac{kt}{2}}}.\frac{{{d}^{2}}x}{d{{t}^{2}}}+\frac{k}{2}{{e}^{\frac{kt}{2}}}\frac{dx}{dt}+\frac{k}{2}{{e}^{\frac{kt}{2}}}\frac{dx}{dt}+\frac{{{k}^{2}}}{4}{{e}^{\frac{kt}{2}}}x=-A{{p}^{2}}\cos (pt+\varepsilon )\]
\[\Rightarrow {{e}^{\frac{kt}{2}}}\left( \frac{{{d}^{2}}x}{d{{t}^{2}}}+k\frac{dx}{dt}+\frac{{{k}^{2}}}{4}x \right)=-{{p}^{2}}{{e}^{\frac{kt}{2}}}x[by(1)]\]
\[\Rightarrow \frac{{{d}^{2}}x}{d{{t}^{2}}}+k\frac{dx}{dt}+\frac{{{k}^{2}}}{4}x=-{{p}^{2}}x\]
\[\Rightarrow \frac{{{d}^{2}}x}{d{{t}^{2}}}+k\frac{dx}{dt}+\left( \frac{{{k}^{2}}}{4}+{{p}^{2}} \right)x=0\]
\[\therefore \frac{{{d}^{2}}x}{d{{t}^{2}}}+k\frac{dx}{dt}+{{n}^{2}}x=0\left[ \because {{n}^{2}}={{p}^{2}}+\frac{1}{4}{{k}^{2}} \right]\]

\[If,y={{x}^{n-1}}{{\log }_{e}}x,show-that\]
\[{{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+\left( 3-2n \right)x\frac{dy}{dx}+{{(n-1)}^{2}}y=0\]
Solution:
\[y={{x}^{n-1}}{{\log }_{e}}x\]
\[\therefore \frac{dy}{dx}=(n-1){{x}^{n-2}}{{\log }_{e}}x+{{x}^{n-1}}.\frac{1}{x}\]
\[\Rightarrow \frac{dy}{dx}=(n-1){{x}^{n-2}}{{\log }_{e}}x+{{x}^{n-2}}\]
\[\Rightarrow \frac{dy}{dx}={{x}^{n-2}}\left[ (n-1){{\log }_{e}}x+1 \right]\]
\[\therefore \frac{{{d}^{2}}y}{d{{x}^{2}}}=(n-2){{x}^{n-3}}.\left[ (n-1){{\log }_{e}}x+1 \right]+{{x}^{n-2}}.(n-1).\frac{1}{x}\]
\[\Rightarrow \frac{{{d}^{2}}y}{d{{x}^{2}}}=(n-2){{x}^{n-3}}.\left[ (n-1){{\log }_{e}}x+1 \right]+(n-1){{x}^{n-3}}\]
\[\Rightarrow {{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}=(n-2){{x}^{n-1}}.\left[ (n-1){{\log }_{e}}x+1 \right]+(n-1){{x}^{n-1}}\]
\[L.H.S.={{x}^{2}}\frac{{{d}^{2}}y}{d{{x}^{2}}}+\left( 3-2n \right)x\frac{dy}{dx}+{{(n-1)}^{2}}y\]
\[=(n-2){{x}^{n-1}}.\left[ (n-1){{\log }_{e}}x+1 \right]+(n-1){{x}^{n-1}}+(3-2n)x.{{x}^{n-2}}\left[ (n-1){{\log }_{e}}x+1 \right]\]
\[+{{(n-1)}^{2}}{{x}^{n-1}}{{\log }_{e}}x\]
\[={{x}^{n-1}}\left[ (n-2)(n-1){{\log }_{e}}x+(n-2)+(n-1)+(3-2n)(n-1){{\log }_{e}}x+(3-2n) \right]\]
\[+{{(n-1)}^{2}}{{x}^{n-1}}{{\log }_{e}}x\]
\[={{x}^{n-1}}\left[ ({{n}^{2}}-3n+2){{\log }_{e}}x+(-2{{n}^{2}}+5n-3){{\log }_{e}}x \right]+{{(n-1)}^{2}}{{x}^{n-1}}{{\log }_{e}}x\]
\[={{x}^{n-1}}{{\log }_{e}}x\left[ {{n}^{2}}-3n+2-2{{n}^{2}}+5n-3+{{n}^{2}}-2n+1 \right]\]
\[={{x}^{n-1}}{{\log }_{e}}x.(0)=0=R.H.S.\]

\[If,p{{v}^{\gamma }}=c,show-that\]
\[{{v}^{2}}\frac{{{d}^{2}}p}{d{{v}^{2}}}=\gamma (\gamma +1)p\]
Solution:
\[p{{v}^{\gamma }}=c\]
\[\therefore {{v}^{\gamma }}\frac{dp}{dv}+p.\gamma {{v}^{\gamma -1}}=0\Rightarrow v\frac{dp}{dv}=-p\gamma [\because v\ne 0]\]
\[\therefore v\frac{{{d}^{2}}p}{d{{v}^{2}}}+\frac{dp}{dv}.1=-\gamma \frac{dp}{dv}\]
\[\Rightarrow v\frac{{{d}^{2}}p}{d{{v}^{2}}}=-(1+\gamma )\frac{dp}{dv}\]
\[\Rightarrow {{v}^{2}}\frac{{{d}^{2}}p}{d{{v}^{2}}}=-(1+\gamma )v\frac{dp}{dv}\]
\[\therefore {{v}^{2}}\frac{{{d}^{2}}p}{d{{v}^{2}}}=\gamma (\gamma +1)p\]

If you want me to solve your 2nd order derivative sum, just comment it, I will be happy to solve it for you and will upload it in the updated post and send it to you. 

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