Ex 5.3 Class 12 Maths Ncert Solutions

It is given that \( 2 x+3 y=\sin x \)
Differentiating both sides w.r.t. \(x \), we get,
\(=\frac{d}{d x}(2 x)+\frac{d}{d x}(3 y)=\frac{d}{d x}(\sin x)\)
\(=2+3 \frac{d y}{d x}=\cos x\)
\(=3 \frac{d y}{d x}=\cos x-2\)
\(=\frac{d y}{d x}=\frac{\cos x-2}{3}\)

It is given that \( 2 x+3 y=\sin y \)
Differentiating both sides w.r.t. \(x \), we get,
\(=\frac{d}{d x}(2 x)+\frac{d}{d x}(3 y)=\frac{d}{d x}(\sin y)\)
\(=2+3 \frac{d y}{d x}=\cos y \frac{d y}{d x}\)
\(=2=(\cos y-3) \frac{d y}{d x}\)
\(=\frac{d y}{d x}=\frac{2}{(\cos y-3)}\)

It is given that \( \mathrm{a} x+\mathrm{b} y^{2}=\cos y \)
Differentiating both sides w.r.t. \(x \), we get,
\(\frac{d}{d x}\left(\mathrm{a} x+\mathrm{b} y^{2}\right)=\frac{d}{d x}(\cos y)\)
\(=\frac{d}{d x}(\mathrm{a} x)+\frac{d}{d x}\left(\mathrm{b}y^{2}\right)=\frac{d}{d x}(\cos y)\)
\(=\mathrm{a}+\mathrm{b} \frac{d}{d x}\left(\mathrm{y}^{2}\right)=\frac{d}{d x}(\cos y)\)
\(=\mathrm{a}+\mathrm{b} \times 2 y \frac{d y}{d x}=-\sin \mathrm{y} \frac{d y}{d x}\)
\(=(2 \mathrm{~b} y+\sin y) \frac{d y}{d x}=-\mathrm{a}\)
\(=\frac{d y}{d x}=\frac{-a}{(2 b y+\sin y)}\)

It is given that \( x y+y^{2}=\tan x+y \)
Differentiating both sides w.r.t. \(x \), we get,
\(\frac{d}{d x}\left(x y+y^{2}\right)=\frac{d}{d x}(\tan x+y)\)
\(=\frac{d}{d x}(x y)+\frac{d}{d x}\left(y^{2}\right)=\frac{d}{d x}(\tan x)+\frac{d y}{d x}\)
\(=\left[y \frac{d y}{d x}(x)+x \frac{d y}{d x}\right]+2 y \frac{d y}{d x}=\sec ^{2}+\frac{d y}{d x}\)
\(=y .1+x \frac{d y}{d x}+2 y \frac{d y}{d x}=\sec ^{2} x+\frac{d y}{d x}\)
\(=(x+2 y-1) \frac{d y}{d x}=\sec 2-y\)
\(=\frac{d y}{d x}=\frac{\sec ^{2} x-y}{(x+2 y-1)}\)

It is given that \( x^{2}+x y+y^{2}=100 \)
Differentiating both sides w.r.t. \(x\), we get,
\(\frac{d}{d x}\left(x^{2}+x y+y^{2}\right)=\frac{d}{d x}(100)\)
\(=\frac{d}{d x}\left(x^{2}\right)+\frac{d}{d x}(x y)+\frac{d}{d x}\left(y^{2}\right)=0\)
\(=2 x+\left[y \frac{d}{d x}(x)+x \frac{d y}{d x}\right]+2 y \frac{d y}{d x}=0\)
\(=2 x+y \cdot 1+x \frac{d y}{d x}+2 y \frac{d y}{d x}=0\)
\(=2 x+y+(x+2 y) \frac{d y}{d x}=0\)
\(=\frac{d y}{d x}=-\frac{2 x+y}{x+2 y}\)

It is given that \( x^{3}+x^{2} y+x y^{2}+y^{3}=81 \)
Differentiating both sides w.r.t. \(x\), we get,
\(\frac{d}{d x}\left(x^{3}+x^{2} y+x y^{2}+y^{3}\right)=\frac{d}{d x}(81)\)
\(=\frac{d}{d x}\left(x^{3}\right)+\frac{d}{d x}\left(x^{2} y\right)+\frac{d}{d x}\left(x y^{2}\right)+\frac{d}{d x}\left(y^{3}\right)=0\)
\(=3 x^{2}+\left[y \frac{d}{d x}\left(x^{2}\right)+x^{2} \frac{d}{d x}\right]+\left[y^{2} \frac{d}{d x}(x)+x \frac{d}{d x}\left(y^{2}\right)\right]+3 y^{2} \frac{d y}{d x}=0\)
\(=3 x^{2}+\left[y .2 x+x^{2} \frac{d y}{d x}\right]+\left[y^{2} \cdot 1+x .2 y \cdot \frac{d y}{d x}\right]+3 y^{2} \frac{d y}{d x}=0\)
\(=\left(x^{2}+2 x y+3 y^{2}\right) \frac{d y}{d x}+\left(3 x^{2}+2 x y+y^{2}\right)=0\)
\(=\frac{d y}{d x}=\frac{-\left(3 x^{2}+2 x y+y^{2}\right)}{\left(x^{2}+2 x y+3 y^{2}\right)}\)

It is given that \( \sin 2 y+\cos x y=\pi \)
Differentiating both sides w.r.t. \(x\), we get,
\(\frac{d}{d x}(\sin 2 y+\cos x y)=\frac{d}{d x}(\pi)\)
\(=2 \sin y \cos y \frac{d y}{d x}-\sin x y\left[y \frac{d}{d x}(x)+x \frac{d y}{d x}\right]\)
\(=2 \sin y \cos y \frac{d y}{d x}-\sin x y\left[y .1+x \frac{d y}{d x}\right]=0\)
\(=2 \sin y \cos y \frac{d y}{d x}-\mathrm{y} \sin x y-x \sin x y \frac{d y}{d x}=0\)
\(=(2 \sin y \cos y-x \sin x y) \frac{d y}{d x}=\mathrm{y} \sin x y\)
\(=(\sin 2 y-x \sin x y) \frac{d y}{d x}=y \sin x y\)
\(=\frac{d y}{d x}=\frac{y \sin x y}{(\sin 2 y-x \sin x y)}\)

It is given that \( \sin 2 x+\cos 2 y=1 \)
Differentiating both sides w.r.t. \( x \), we get,
\(\frac{d}{d x}(\sin 2 x+\cos 2 y)=\frac{d}{d x} \ldots(1)\)
\(=\frac{d}{d x}(\sin 2 x)+\frac{d}{d x}(\cos 2 x)=0\)
\(=2 \sin x \cdot \frac{d}{d x}(\sin x)+2 \cos y \cdot \frac{d}{d x}(\cos y)=0\)
\(=2 \sin x \cos x+2 \cos y(-\sin y) \cdot \frac{d y}{d x}=0\)
\(=\sin 2 x-\sin 2 y \frac{d y}{d x}=0\)
\(=\frac{d y}{d x}=\frac{\sin 2 x}{\sin 2 y}\)

Let \( x=\tan \mathrm{A} \)
then, \( \mathrm{A}=\tan ^{-1} x \)
\(=\frac{d A}{d x}=\frac{1}{1+x^{2}}\)
\(y=\sin ^{-1}\left(\frac{2 \tan A}{1+\tan ^{2} A}\right)\)
also, we know \( \left[\sin 2 A=\frac{2 \tan A}{1+\tan ^{2} A}\right] \)
And \( =y=\sin ^{-1}(\sin 2 \mathrm{~A}) \)
\( =y=2 \mathrm{~A} \)
\( =\frac{d y}{d x}=2 \frac{d A}{d x} \quad \)[by chain rule]
\(=\frac{d y}{d x}=\frac{2}{1+x^{2}}\)

It is given that:
\(y=\tan ^{-1}\left(\frac{3 x-x^{3}}{1-3 x^{2}}\right)\)
Assumption: Let \( x=\tan \theta \), putting it in \(y \), we get,
\( y=\tan ^{-1}\left(\frac{3 \tan \theta-\tan ^{3} \theta}{1-3 \tan ^{2} \theta}\right) \)
we know by the formula that, \( \tan 3 x=\frac{3 \tan x-\tan ^{3} x}{1-3 \tan ^{2} x} \)
Putting this in \( y \), we get, \( y=\tan ^{-1}(\tan 3 \theta) \)
\(y=3\left(\tan ^{-1} x\right)\)
Differentiating both sides, we get, \( \frac{d y}{d x}=\frac{3}{1+x^{2}} \)

It is given that,
\(y=\cos ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right)\)
\( =\cos y=\frac{1-x^{2}}{1+x^{2}} \)
\( =\frac{1-\tan ^{2} \frac{y}{2}}{1+\tan ^{2} \frac{y}{2}}=\frac{1-x^{2}}{1+x^{2}} \)
On comparing both sides, we get,
\(\tan \frac{y}{2}=x\)
Now, differentiating both sides, we get,
\(\operatorname{Sec}^{2}\left(\frac{y}{2}\right) \cdot \frac{d}{d x}\left(\frac{y}{2}\right)=\frac{d}{d x}(x)\)
\(=\sec ^{2} \frac{y}{2} \times \frac{1}{2} \frac{d y}{d x}=1\)
\(=\frac{d y}{d x}=\frac{2}{\sec ^{2} \frac{y}{2}}\)
\(=\frac{d y}{d x}=\frac{2}{1+\tan ^{2} \frac{y}{2}}\)
\(=\frac{d y}{d x}=\frac{2}{1+x^{2}}\)

It is given that \( y=\sin ^{-1}\left(\frac{1-x^{2}}{1+x^{2}}\right) \)
\(=\sin y=\frac{1-x^{2}}{1+x^{2}}\)
\(=\left(1+x^{2}\right) \sin y=1-x^{2}\)
\(=(1+\sin y) x^{2}=1-\sin y\)
\(=x^{2}=\frac{1-\sin y}{1+\sin y}\)
Now, we can change the numerator and the denominator,
\(1=\sin^2 \frac{y}{2}+\cos^2 \frac{y}{2}\)
We know that we can write, and
\( \operatorname{Sin} y=2 \sin \frac{y}{2} . \operatorname{Cos} \frac{y}{2} \)
Therefore, by applying the formula: \( (a+b)^{2}=a^{2}+b^{2}+2 a b \) and \( (a-b)^{2} \) \( =a^{2}+b^{2}-2 a b \), we get,
\( =x^{2}=\frac{\left(\cos \frac{y}{2}-\sin \frac{y}{2}\right)^{2}}{\left(\cos \frac{y}{2}+\sin \frac{y}{2}\right)^{2}} \)
\( =x=\frac{\cos \frac{y}{2}-\sin \frac{y}{2}}{\cos \frac{y}{2}+\sin \frac{y}{2}} \)
Dividing the numerator and denominator by \( \cos (\frac{y}{2}) \), we get, \( =x \frac{1-\tan \frac{y}{2}}{1+\tan \frac{y}{2}} \)
Now, we know that:
\( \tan (\mathrm{A}-\mathrm{B})=\frac{\tan A-\tan B}{1+\tan A \cdot \tan B} \)
\( =x \tan \left(\frac{\pi}{4}-\frac{y}{2}\right) \)
Now, differentiating both sides, we get,
\(\frac{d}{d x}(x)=\frac{d}{d x}\left(\tan \left(\frac{\pi}{4}-\frac{y}{2}\right)\right)\)
\(=1=\sec ^{2}\left(\frac{\pi}{2}-\frac{y}{2}\right) \times \frac{d}{d x}\left(\frac{\pi}{4}-\frac{y}{2}\right)\)
\(=1=\left[1+\tan ^{2}\left(\frac{\pi}{4}-\frac{y}{2}\right)\right] \cdot\left(-\frac{1}{2} \frac{d y}{d x}\right)\)
\(=1=\left[1+x^{2}\right] \cdot\left(-\frac{1}{2} \frac{d y}{d x}\right)\)
\(=\frac{d y}{d x}=\frac{-2}{1+x^{2}}\)

It is given that \( y=\cos ^{-1}\left(\frac{2 x}{1+x^{2}}\right) \)
\(=\cos y=\frac{2 x}{1+x^{2}}\)
Differentiating both sides w.r.t. \(x \), we get,
\(-\operatorname{Sin} y \frac{d y}{d x}=\frac{\left(1+x^{2}\right) \cdot \frac{d}{d x}(2 x)-2 x \cdot \frac{d}{d x}\left(1+x^{2}\right)}{\left(1+x^{2}\right)^{2}}\)
\(=\sqrt{1-\cos ^{2} y} \frac{d y}{d x}=\frac{\left(1+x^{2}\right) \times 2-2 x \cdot 2 x}{\left(1+x^{2}\right)^{2}}\)
\(=\sqrt{1-\left(\frac{2 x}{1+x^{2}}\right)^{2}} \frac{d y}{d x}=\left[\frac{\left(1-x^{2}\right)}{\left(1+x^{2}\right)}\right]\)
\(=\sqrt{\frac{\left(1-x^{2}\right)^{2}-4 x^{2}}{\left(1+x^{2}\right)^{2}}} \frac{d y}{d x}=\frac{-2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}\)
\(=\sqrt{\frac{\left(1-x^{2}\right)^{2}}{\left(1+x^{2}\right)^{2}} \frac{d y}{d x}}=\frac{-2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}\)
\(=\frac{1-x^{2}}{1+x^{2}} \frac{d y}{d x}=\frac{-2\left(1-x^{2}\right)}{\left(1+x^{2}\right)^{2}}\)
\(=\frac{d y}{d x}=\frac{-2}{1+x^{2}}\)

It is given that \( y=\sin ^{-1}\left(2 x \sqrt{1-x^{2}}\right) \)
\(=\sin y=2 x \sqrt{1-x^{2}}\)
Differentiating both sides w.r.t. \(x \), we get,
\(\operatorname{Cos} y \frac{d y}{d x}=2\left[x \frac{d}{d x}\left(\sqrt{1-x^{2}}\right)+\sqrt{1-x^{2}} \frac{d y}{d x}\right]\)
\(=\sqrt{1-\sin^{2} y} \frac{d y}{d x}=2\left[\frac{x}{2} \cdot \frac{-2 x}{\sqrt{1-x^{2}}}+\sqrt{1+x^{2}}\right]\)
\(=\sqrt{1-\left(2 x \sqrt{1-x^{2}}\right)^{2}} \frac{d y}{d x}=2\left[\frac{-x^{2}+1-x^{2}}{\sqrt{1-x^{2}}}\right]\)
\(=\sqrt{1-4 x^{2}\left(1-x^{2}\right)} \frac{d y}{d x}=2\left[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\right]\)
\(=(1-2 x^2) \frac{d y}{d x}=2\left[\frac{1-2 x^{2}}{\sqrt{1-x^{2}}}\right]\)
\(=\frac{d y}{d x}=\frac{2}{\sqrt{1-x^{2}}}\)

It is given that \( y=\sec ^{-1}\left(\frac{1}{2 x^{2}+1}\right) \)
\(=\sec y=\frac{1}{2 x^{2}+1}\)
\(=\cos y=2 x^{2}+1\)
\(=2 x^{2}=1+\cos y\)
\(=2 x^{2}=2 \cos^2 \frac{y}{2}\)
\(=x=\cos \frac{y}{2}\)
Differentiating w.r.t. \( x \), we get,
\(\frac{d}{d x}(x)=\frac{d}{d x}\left(\cos \frac{y}{2}\right)\)
\(=1=-\sin \frac{y}{2} \cdot \frac{d}{d x}\left(\frac{y}{2}\right)\)
\(=\frac{-1}{\sin \frac{y}{2}}=\frac{1}{2} \frac{d y}{d x}\)
\(=\frac{d y}{d x}=\frac{-2}{\sin \frac{y}{2}}=\frac{-2}{\sqrt{1-\cos ^{2} \frac{y}{2}}}\)
\(=\frac{d y}{d x}=\frac{-2}{\sqrt{1-x^{2}}}\)