80258
If \(y=\left(\tan ^{-1} x\right)^{2}\), then \(\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x\left(x^{2}+1\right) \frac{d y}{d x}=\)
1 4
2 2
3 1
4 0
Explanation:
(B) : Given \(\frac{d y}{d x}=\frac{2 \tan ^{-1} x}{1+x^{2}}\) \(\frac{d^{2} y}{d x^{2}}=\frac{\frac{2}{\left(1+x^{2}\right)} \times\left(1+x^{2}\right)-2 \tan ^{-1} x \cdot 2 x}{\left(1+x^{2}\right)^{2}}\) \(\frac{d^{2} y}{d x^{2}}=\frac{2-4 x \tan ^{-1} x}{\left(1+x^{2}\right)^{2}}\) \(\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}=2-4 x \tan ^{-1} x \tag{ii}\) Taking equation (i), \(\frac{d y}{d x}=\frac{2 \tan ^{-1} x}{1+x^{2}}\) \(\left(1+x^{2}\right) \cdot \frac{d y}{d x}=2 \tan ^{-1} x \tag{iii}\) Therefore, \(\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x\left(x^{2}+1\right) \frac{d y}{d x}\) Putting the value equation (i) and (ii) in above equation, \(=2-4 x \tan ^{-1} x+2 x \times 2 \tan ^{-1} x\) \(=2-4 x \tan ^{-1} x+4 x \tan ^{-1} x=2\)
MHT CET-2018
Limits, Continuity and Differentiability
80259
If \(x=e^{\theta}(\sin \theta-\cos \theta), y=e^{\theta}(\sin \theta+\cos \theta)\), then then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) at \(\theta=\frac{\pi}{4}\) is
80258
If \(y=\left(\tan ^{-1} x\right)^{2}\), then \(\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x\left(x^{2}+1\right) \frac{d y}{d x}=\)
1 4
2 2
3 1
4 0
Explanation:
(B) : Given \(\frac{d y}{d x}=\frac{2 \tan ^{-1} x}{1+x^{2}}\) \(\frac{d^{2} y}{d x^{2}}=\frac{\frac{2}{\left(1+x^{2}\right)} \times\left(1+x^{2}\right)-2 \tan ^{-1} x \cdot 2 x}{\left(1+x^{2}\right)^{2}}\) \(\frac{d^{2} y}{d x^{2}}=\frac{2-4 x \tan ^{-1} x}{\left(1+x^{2}\right)^{2}}\) \(\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}=2-4 x \tan ^{-1} x \tag{ii}\) Taking equation (i), \(\frac{d y}{d x}=\frac{2 \tan ^{-1} x}{1+x^{2}}\) \(\left(1+x^{2}\right) \cdot \frac{d y}{d x}=2 \tan ^{-1} x \tag{iii}\) Therefore, \(\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x\left(x^{2}+1\right) \frac{d y}{d x}\) Putting the value equation (i) and (ii) in above equation, \(=2-4 x \tan ^{-1} x+2 x \times 2 \tan ^{-1} x\) \(=2-4 x \tan ^{-1} x+4 x \tan ^{-1} x=2\)
MHT CET-2018
Limits, Continuity and Differentiability
80259
If \(x=e^{\theta}(\sin \theta-\cos \theta), y=e^{\theta}(\sin \theta+\cos \theta)\), then then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) at \(\theta=\frac{\pi}{4}\) is
80258
If \(y=\left(\tan ^{-1} x\right)^{2}\), then \(\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x\left(x^{2}+1\right) \frac{d y}{d x}=\)
1 4
2 2
3 1
4 0
Explanation:
(B) : Given \(\frac{d y}{d x}=\frac{2 \tan ^{-1} x}{1+x^{2}}\) \(\frac{d^{2} y}{d x^{2}}=\frac{\frac{2}{\left(1+x^{2}\right)} \times\left(1+x^{2}\right)-2 \tan ^{-1} x \cdot 2 x}{\left(1+x^{2}\right)^{2}}\) \(\frac{d^{2} y}{d x^{2}}=\frac{2-4 x \tan ^{-1} x}{\left(1+x^{2}\right)^{2}}\) \(\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}=2-4 x \tan ^{-1} x \tag{ii}\) Taking equation (i), \(\frac{d y}{d x}=\frac{2 \tan ^{-1} x}{1+x^{2}}\) \(\left(1+x^{2}\right) \cdot \frac{d y}{d x}=2 \tan ^{-1} x \tag{iii}\) Therefore, \(\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x\left(x^{2}+1\right) \frac{d y}{d x}\) Putting the value equation (i) and (ii) in above equation, \(=2-4 x \tan ^{-1} x+2 x \times 2 \tan ^{-1} x\) \(=2-4 x \tan ^{-1} x+4 x \tan ^{-1} x=2\)
MHT CET-2018
Limits, Continuity and Differentiability
80259
If \(x=e^{\theta}(\sin \theta-\cos \theta), y=e^{\theta}(\sin \theta+\cos \theta)\), then then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) at \(\theta=\frac{\pi}{4}\) is
NEET Test Series from KOTA - 10 Papers In MS WORD
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Limits, Continuity and Differentiability
80258
If \(y=\left(\tan ^{-1} x\right)^{2}\), then \(\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x\left(x^{2}+1\right) \frac{d y}{d x}=\)
1 4
2 2
3 1
4 0
Explanation:
(B) : Given \(\frac{d y}{d x}=\frac{2 \tan ^{-1} x}{1+x^{2}}\) \(\frac{d^{2} y}{d x^{2}}=\frac{\frac{2}{\left(1+x^{2}\right)} \times\left(1+x^{2}\right)-2 \tan ^{-1} x \cdot 2 x}{\left(1+x^{2}\right)^{2}}\) \(\frac{d^{2} y}{d x^{2}}=\frac{2-4 x \tan ^{-1} x}{\left(1+x^{2}\right)^{2}}\) \(\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}=2-4 x \tan ^{-1} x \tag{ii}\) Taking equation (i), \(\frac{d y}{d x}=\frac{2 \tan ^{-1} x}{1+x^{2}}\) \(\left(1+x^{2}\right) \cdot \frac{d y}{d x}=2 \tan ^{-1} x \tag{iii}\) Therefore, \(\left(x^{2}+1\right)^{2} \frac{d^{2} y}{d x^{2}}+2 x\left(x^{2}+1\right) \frac{d y}{d x}\) Putting the value equation (i) and (ii) in above equation, \(=2-4 x \tan ^{-1} x+2 x \times 2 \tan ^{-1} x\) \(=2-4 x \tan ^{-1} x+4 x \tan ^{-1} x=2\)
MHT CET-2018
Limits, Continuity and Differentiability
80259
If \(x=e^{\theta}(\sin \theta-\cos \theta), y=e^{\theta}(\sin \theta+\cos \theta)\), then then \(\frac{\mathrm{dy}}{\mathrm{dx}}\) at \(\theta=\frac{\pi}{4}\) is
