Ex 5.7 Class 12 Maths Ncert Solutions

Let us take \( y=x^{2}+3 x+2 \)
Now,
\(\frac{d y}{d x}=\frac{d\left(x^{2}\right)}{d x}+\frac{d(3 x)}{d x}+\frac{d(2)}{d x}\)
\(=2 x+3\)
Therefore,
\(\frac{d^{2} y}{d x^{2}}=\frac{d(2 x+3)}{d x}=\frac{d(2 x)}{d x}+\frac{d(3)}{d x}\)
\(=2+0\)
\(=2\)

Let us take \( y=x^{20} \)
Now,
\(\frac{d y}{d x}=\frac{d\left(x^{20}\right)}{d x}\)
\(=20 x^{19}\)
Therefore,
\(\frac{d^{2} y}{d x^{2}}=\frac{d\left(20 x^{19}\right)}{d x}=20 \frac{d\left(x^{19}\right)}{d x}\)
\(=20 \times 19 \times x^{18}\)
\(=380 x^{18}\)

Let us take \( y=x \cdot \cos x \)
Now,
\(\frac{d y}{d x}=\frac{d(x \cos x)}{d x}\)
\(=\cos x \frac{d(x)}{d x}+x \frac{d(\cos x)}{d x}\)
\(=\cos x .1+x(-\sin x)\)
\(=\cos x-x \sin x\)
Therefore,
\(\frac{d^{2} y}{d x^{2}}=\frac{d(\cos x-x \sin x)}{d x}\)
\(=\frac{d(\cos x)}{d x}-\frac{d(x \sin x)}{d x}\)
\(=-\sin x-\left[\sin x \cdot \frac{d(x)}{d x}+x \cdot \frac{d(\sin x)}{d x}\right]\)
\(=-\sin x-(\sin x+x \cos x)\)
\(=-(x \cos x+2 \sin x)\)

Let us take \( y=\log x \)
Now,
\(\frac{d y}{d x}=\frac{d(\log x)}{d x}=\frac{1}{x}\)
Therefore,
\(\frac{d^{2} y}{d x^{2}}=\frac{d\left(\frac{1}{x}\right)}{d x}=\left(-\frac{1}{x^{2}}\right)\)

Let us take \( y=x^{3} \log x \)
Now,
\(\frac{d y}{d x}=\frac{d\left(x^{3} \log x\right)}{d x}\)
\(=\log x \cdot \frac{d\left(x^{3}\right)}{d x}+x^{3} \cdot \frac{d(\log x)}{d x}\)
\(= \log x \cdot 3 x^{2}+x^{3} \cdot \frac{1 }{ x }\)
\(=\log x \cdot 3 x^{2}+x^{2}\)
\(=x^{2}(1+3 \log x)\)
Therefore,
\(\frac{d^{2} y}{d x^{2}}=\frac{d\left[x^{2}(1+3 \log x)\right]}{d x}\)
\(=(1+3 \log x) \cdot \frac{d\left(x^{2}\right)}{d x}+x^{2} \frac{d(1+3 \log x)}{d x}\)
\(=(1+3 \log x) \cdot 2 x+x^{2} \cdot \frac{3}{x}\)
\(=2 x+6 x \log x+3 x\)
\(=5 x+6 x \log x\)
\(=x(5+6 \log x)\)

Let us take \( y=\mathrm{e}^{x} \sin 5 x \)
Now,
\(\frac{d y}{d x}=\frac{d\left(e^{x} \sin 5 x\right)}{d x}\)
\(=\sin 5 x \cdot \frac{d\left(e^{x}\right)}{d x}+e^{x} \cdot \frac{d(\sin 5 x)}{d x}\)
\(=\sin 5 x \cdot e^{x}+e^{x} \cdot \cos 5 x \cdot \frac{d(5 x)}{d x}\)
\(=\mathrm{e}^{x} \sin 5 x+\mathrm{e}^{x} \cos 5 x \cdot 5\)
\(=\mathrm{e}^{x}(\sin 5 x+5 \cos 5 x)\)
\(\frac{d^{2} y}{d x^{2}}=\frac{d\left[e^{x}(\sin 5 x+5 \cos 5 x)\right]}{d x}\)
\(=(\sin 5 x+5 \cos 5 x) \cdot \frac{d\left(e^{x}\right)}{d x}+e^{x} \cdot \frac{d(\sin 5 x+5 \cos 5 x)}{d x}\)
\(=(\sin 5 x+5 \cos 5 x) e^{x}+e^{x}\left[\cos 5 x \cdot \frac{d(5 x)}{d x}+5(-\sin 5 x) \cdot \frac{d(5 x)}{d x}\right]\)
\(=\mathrm{e}^{x}(\sin 5 x+5 \cos 5 x)+\mathrm{e}^{x}(5 \cos 5 x-25 \sin 5 x)\)
\(=\mathrm{e}^{x}(10 \cos 5 x-24 \sin 5 x)\)
\(=2 \mathrm{e}^{x}(5 \cos 5 x-12 \sin 5 x)\)

Let us take \( y=e^{6 x} \cos 3 x \)
Now,
\(\frac{d y}{d x}=\frac{d\left(e^{6 x} \cos 3 x\right)}{d x}\)
\(=\cos 3 x \cdot \frac{d\left(e^{6 x}\right)}{d x}+e^{6 x} \frac{d(\cos 3 x)}{d x}\)
\(=\cos 3 x \cdot e^{6 x} \cdot \frac{d(6 x)}{d x}+e^{6 x} \cdot(-\sin 3 x) \cdot \frac{d(3 x)}{d x}\)
\(=6 \mathrm{e} 6 x \cos 3 x-3 \mathrm{e} 6 x \sin 3 x\)
\(\frac{d^{2} y}{d x^{2}}=\frac{d\left[6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x\right]}{d x}\)
\(=6 \cdot \frac{d\left(e^{6 x} \cos 3 x\right)}{d x}-3 \cdot \frac{d\left(e^{6 x} \sin 3 x\right)}{d x}\)
\(=6 \cdot\left[6 e^{6 x} \cos 3 x-3 e^{6 x} \sin 3 x\right]-3\left[\sin 3 x \cdot \frac{d\left(e^{6 x}\right)}{d x}+e^{6 x} \frac{d(\sin 3 x)}{d x}\right]\)
\(=36 \mathrm{e}^{6 x} \cos 3 x-18 \mathrm{e}^{6 x} \sin 3 x-3\left[\sin 3 x \cdot \mathrm{e}^{6 x} \cdot 6+\mathrm{e}^{6 x} \cdot \cos 3 x \cdot 3\right]\)
\(=36 \mathrm{e}^{6 x} \cos 3 x-18 \mathrm{e}^{6 x} \sin 3 x-18 \mathrm{e}^{6 x} \sin 3 x-9 \mathrm{e}^{6 x} \cos 3 x\)
\(=27 \mathrm{e}^{6 x} \cos 3 x-36 \mathrm{e}^{6 x} \sin 3 x\)
\(=9 \mathrm{e}^{6 x}(3 \cos 3 x-4 \sin 3 x)\)

Let us take \( y=\tan ^{-1} x \) Now,
\(\frac{d y}{d x}=\frac{d\left(\tan ^{-1}\right)}{d x}=\frac{1}{1+x^{2}}\)
\(\frac{d^{2} y}{d x^{2}}=\frac{d\left[\frac{1}{1+x^{2}}\right]}{d x}\)
\(=\frac{d\left(1+x^{2}\right)^{-1}}{d x}=(-1) \cdot\left(1+x^{2}\right) \cdot \frac{d\left(1+x^{2}\right)}{d x}\)
\(=\frac{1}{\left(1+x^{2}\right)^{2}} \times 2 x=\frac{-2 x}{\left(1+x^{2}\right)^{2}}\)

Let us take \( y=\log (\log x) \)
Now,
\(\frac{d y}{d x}=\frac{d[\log (\log x)]}{d x}\)
\(=\frac{1}{\log x} \cdot \frac{d(\log x)}{d x}=\frac{1}{x \log x}\)
\(=(x \log x)^{-1}\)
\(\frac{d^{2} y}{d x^{2}}=\frac{d(x \log x)^{-1}}{d x}\)
\(=(-1) \cdot(x \log x)^{-2} \cdot \frac{d(x \log x)}{d x}\)
\(=\frac{-1}{(x \log x)^{2}} \cdot\left[\log x \cdot \frac{d(x)}{d x}+x \cdot \frac{d(\log x)}{d x}\right]\)
\(=\frac{-1}{(x \log x)^{2}} \cdot\left[\log x \cdot 1+x \cdot \frac{1}{x}\right]\)
\(=\frac{-(1+\log x)}{(x \log x)^{2}}\)

Let us take \( y=\sin (\log x) \)
Now,
\(\frac{d y}{d x}=\frac{d[\sin (\log x)]}{d x}\)
\(=\cos (\log x) \cdot \frac{d(\log x)}{d x}\)
\(=\frac{\cos (\log x)}{x}\)
Then
\(\frac{d^{2} y}{d x^{2}}=\frac{d\left(\frac{\cos (\log x)}{x}\right)}{d x}\)
\(=\frac{x \cdot \frac{d[\cos (\log x)]}{d x}-\cos (\log x) \cdot \frac{d(x)}{d x}}{x^{2}}\)
\(=\frac{x \cdot\left[-\sin (\log x) \cdot \frac{d(\log x)}{d x}\right]-\cos (\log x) \cdot 1}{x^{2}}\)
\(=\frac{-x \sin (\log x) \cdot \frac{1}{x} \cdot \cos (\log x)}{x^{2}}\)
\(=\frac{-\sin (\log x)+\cos (\log x)}{x^{2}}\)

It is given that \( y=5 \cos x-3 \sin x \)
Now, on differentiating we get,
\(\frac{d y}{d x}=\frac{d[5 \cos x-3 \sin x]}{d x}\)
\(=\frac{d(5 \cos x)}{d x}-\frac{d(3 \sin x)}{d x}\)
\(=\frac{5 d(\cos 5 x)}{d x}-\frac{3 d(\sin x)}{d x}\)
\(=5(-\sin x)-3(\cos x)\)
\(=-(5 \sin x+\cos x)\)
Then,
\(\frac{d^{2} y}{d x^{2}}=\frac{d(-(5 \sin x+\cos x))}{d x}\)
\(=-\left[5 \cdot \frac{d(\sin x)}{d x}+3 \cdot \frac{d(\cos x)}{d x}\right]\)
\(=-[5 \cos x+3(-\sin x)]\)
\(=-[5 \cos x-3 \sin x]\)
\(=-y\)
Therefore,
\(\frac{d^{2} y}{d x^{2}}+y=0\)
Hence Proved.

It is given that \( y=\cos ^{-1} x \)
Now,
\(\frac{d y}{d x}=\frac{d\left(\cos ^{-1}\right)}{d x}=\frac{-1}{\sqrt{1-x^{2}}}=-\left(1-x^{2}\right)^{-\frac{1}{2}}\)
Therefore,
\(\frac{d^{2} y}{d x^{2}}=\frac{d\left(-\left(1-x^{2}\right)^{-\frac{1}{2}}\right)}{d x}\)
\(=-\left(-\frac{1}{2}\right) \cdot\left(1-x^{2}\right)^{-\frac{3}{2}} \cdot \frac{d\left(1-x^{2}\right)}{d x}\)
\(=\frac{1}{2{\sqrt{1-x^{2}}}^{3}} \times(-2 x)\)
\(\frac{d^{2} y}{d x^{2}}=\frac{-x}{{\sqrt{\left(1-x^{2}\right)}}^{3}}\ldots(1)\)
Now it is given that \( y=\cos ^{-1} x \)
\(\Rightarrow x=\cos y\)
Now putting the value of \( x \) in equation (1), we get
\(\frac{d^{2} y}{d x^{2}}=\frac{-\cos y}{{\sqrt{1-\cos ^{2} y}}^{3}}\)
\(=\frac{-\cos y}{{\sqrt{\sin ^{2} y}}^{3}}\)
\(=\frac{-\cos y}{(\sin y)^{3}}=\frac{-\cos y}{\sin y} \cdot \frac{1}{\sin ^{2} y}\)
\(=\frac{d^{2} y}{d x^{2}}=-\cot y \cdot \operatorname{cosec}^2 y\)

It is given that \( y=3 \cos (\log x)+4 \sin (\log x) \)
Now, on differentiating we get,
\(\frac{d y}{d x}=\frac{d(3 \cos (\log x))+4 \sin (\log x))}{d x}\)
\(=3 \cdot \frac{d(\cos (\log x))}{d x}+4 \cdot \frac{d(\sin (\log x))}{d x}\)
\(=3 \cdot\left[-\sin (\log x) \cdot \frac{d(\log x)}{d x}\right]+4 \cdot\left[\cos (\log x) \cdot \frac{d(\log x)}{d x}\right]\)
\(=\frac{d y}{d x}=\frac{-3 \sin (\log x)}{x}+\frac{4 \cos (\log x)}{x}=\frac{4 \cos (\log x)-3 \sin (\log x)}{x}\)
Again differentiating we get,
\(\frac{d^{2} y}{d x^{2}}=\frac{d\left(\frac{4 \cos (\log x)-3 \sin (\log x)}{x}\right)}{d x}\)
\(=\frac{x\{4 \cos (\log x)-3 \sin (\log x)\}^{\prime}-\{4 \cos (\log x)-3 \sin (\log x)\}(x) \prime}{x^{2}}\)
\(=\frac{x\left[-4 \sin (\log x) \cdot(\log x)^{\prime}-3 \cos (\log x) \cdot(\log x) ^\prime\right]-4 \cos (\log x)+3 \sin (\log x)}{x^{2}}\)
\(=\frac{-4 \sin (\log x)-3 \cos (\log x)-4 \cos (\log x)+3 \sin (\log x)}{x^{2}}\)
\(=\frac{-\sin (\log x)-7 \cos (\log x)}{x^{2}}\)
Therefore,
\(x^{2} y_{2}+x y_{1}+y\)
\(=x^{2}\left(\frac{-\sin (\log x)-7 \cos (\log x)}{x^{2}}\right)+x\left(\frac{4 \cos (\log x)-3 \sin (\log x)}{x}\right)+3 \cos (\log x)+4 \sin (\log x)\)
\(=-\sin (\log x)-7 \cos (\log x)+4 \cos (\log x)-3 \sin (\log x)+3 \cos (\log x)+ 4 \sin (\log x)\)
\(=0\)
So, \( x^{2} y_{2}+xy_{1}+y=0 \)
Hence Proved

According to given equation, we have,
\(y=\mathrm{Ae}^{\mathrm{mx}}+\mathrm{Be}^{\mathrm{nx}}\)
\(\text { Then, } \frac{d y}{d x}=\frac{d\left(A e^{m x}+B e^{n x}\right)}{d x}\)
\(=\text { A. } \frac{d\left(e^{m x}\right)}{d x}+B \cdot \frac{d\left(e^{n x}\right)}{d x}\)
\(=\mathrm{A} \cdot \mathrm{e}^{m x} \frac{d(m x)}{d x}+B \cdot e^{n x} \frac{d(n x)}{d x}\)
\(=\mathrm{Ame}^{\mathrm{mx}}+\mathrm{Bne}^{\mathrm{nx}}\)
Now, on again differentiating we get,
\(\frac{d^{2} y}{d x^{2}}=\frac{d\left(A m e^{m x}+B n e^{n x}\right)}{d x}\)
\(=\mathrm{Am} \cdot \frac{d\left(e^{m x}\right)}{d x}+B n \cdot \frac{d\left(e^{n x}\right)}{d x}\)
\(=\mathrm{Am} \cdot e^{m x} \frac{d(m x)}{d x}+B n \cdot e^{n x} \frac{d(n x)}{d x}\)
\(=\mathrm{Am}^{2} \mathrm{e}^{\mathrm{mx}}+\mathrm{Bn}^{2} \mathrm{e}^{\mathrm{nx}}\)
\(\therefore \frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y\)
\(=\mathrm{Am}^{2} \mathrm{e}^{\mathrm{mx}}+\mathrm{Bn}^{2} \mathrm{e}^{\mathrm{nx}}-(\mathrm{m}+\mathrm{n})\left(\mathrm{Ame}^{\mathrm{mx}}+\mathrm{Bne}^{\mathrm{nx}}\right)+\mathrm{mn}\left(\mathrm{Ae}^{\mathrm{mx}}+\mathrm{Be}^{\mathrm{nx}}\right)\)
\(=\mathrm{Am}^{2} \mathrm{e}^{\mathrm{mx}}+\mathrm{Bn}^{2} \mathrm{e}^{\mathrm{nx}}-\mathrm{Am}^{2} \mathrm{e}^{\mathrm{mx}}-\mathrm{Bmne}^{\mathrm{nx}}-\mathrm{Amne}^{\mathrm{mx}}-\mathrm{Bn}^{2} \mathrm{e}^{\mathrm{ex}}+\mathrm{Amne}^{\mathrm{mx}}+ \mathrm{Bmne}^{\mathrm{nx}}\)
\(=0\)
\(=\frac{d^{2} y}{d x^{2}}-(m+n) \frac{d y}{d x}+m n y=0\)
Hence Proved

According to given equation, we have,
\(y=500 \mathrm{e}^{7 x}+600 \mathrm{e}^{-7 x}\)
\(\frac{d y}{d x}=\frac{d\left(500 e^{7 x}+600 e^{-7 x}\right)}{d x}\)
\(=500 \cdot \frac{d\left(e^{7 x}\right)}{d x}+600 \cdot \frac{d(-7 x)}{d x}\)
\(=500 \cdot e^{7 x} \frac{d(7 x)}{d x}+600 \cdot e^{-7 x} \frac{d(-7 x)}{d x}\)
\(=3500 \mathrm{e}^{7 x}-4200 \mathrm{e}^{-7 x}\)
Now, on again differentiating we get,
\(\frac{d^{2} y}{d x^{2}}=\frac{d\left(3500 e^{7 x}-4200 e^{-7 x}\right)}{d x}\)
\(=3500 \cdot \frac{d\left(e^{7 x}\right)}{d x}-4200 \frac{d\left(e^{-7 x}\right)}{d x}\)
\(=3500 e^{7 x} \frac{d(7 x)}{d x}-42500 e^{-7 x} \frac{d}{d x(-7 x)}\)
\(=7 \times 3500 \cdot \mathrm{e}^{7 x}+7 \times 4200 \cdot \mathrm{e}^{-7 x}\)
\(=49 \times 500 \mathrm{e}^{7 x}+49 \times 600 \mathrm{e}^{-7 x}\)
\(=49\left(500 \mathrm{e}^{7 x}+600 \mathrm{e}^{-7 x}\right)\)
\(=49 y\)
\(\therefore \frac{d^{2} y}{d x^{2}}=49 y\)
Hence Proved

It is given that
\( e^{y}(x+1)=1 \)
\(=\mathrm{e}^{y}=\frac{1}{x+1}\)
Now, taking logarithm on both the sides we get,
\(y=\log \frac{1}{x+1}\)
On differentiating both sides, we get,
\(\frac{d y}{d x}=(x+1) \frac{d\left(\frac{1}{x+1}\right)}{d x}\)
\(=(x+1) \cdot \frac{-1}{(x+1)^{2}}=\frac{-1}{x+1}\)
Again, on differentiating we get,
\(\therefore \frac{d^{2} y}{d x^{2}}=-\frac{d\left(\frac{1}{x+1}\right)}{d x}\)
\(=-\left(\frac{d^{2} y}{d x^{2}}\right)=\frac{1}{(x+1)^{2}}\)
\(=\frac{d^{2} y}{d x^{2}}=\frac{1}{(x+1)^{2}}\)
\(=\frac{d^{2} y}{d x^{2}}=\left(\frac{d y}{d x}\right)^{2}\)
Hence Proved

It is given that
\(y=\left(\tan ^{-1} x\right)^{2}\)
On differentiating we get,
\(\frac{d y}{d x}=\frac{d\left[\left(\tan ^{-1} x\right)^{2}\right]}{d x}\)
\(=2 \tan ^{-1} x \frac{d\left[\tan ^{-1} x\right]}{d x}\)
\(=2 \tan ^{-1} x \frac{1}{1+x^{2}}\)
\(=\left(1+x^{2}\right) \frac{d y}{d x}=2 \tan ^{-1} x\)
Again differentiating, we get,
\(\left(1+x^{2}\right) \frac{d^{2} y}{d x^{2}}+2 x \frac{d y}{d x}=2\left(\frac{1}{1+x^{2}}\right)\)
\(=\left(1+x^{2}\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x\left(1+x^{2}\right) \frac{d y}{d x}=2\)
So, \( \left(1+x^{2}\right) 2 y^{2}+2 x\left(1+x^{2}\right) y_{1}=2 \)
where, \( y_{1}=\frac{d y}{d x} \) and \( y_{2}=\frac{d^{2} y}{d x^{2}} \)
Hence Proved