Differentiation of Function
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Limits, Continuity and Differentiability

80262 If \(x=\sqrt{\mathbf{a}^{\sin ^{-1} t}}, y=\sqrt{\mathbf{a}^{\cos ^{-1} t}}\), then \(\frac{d y}{d x}=\)

1 \(\frac{-x}{y}\)
2 \(\frac{y}{x}\)
3 \(\frac{-y}{x}\)
4 \(\frac{x}{y}\)
MHT CET-2019
Limits, Continuity and Differentiability

80264 If \(x=\sin \theta, y=\sin ^{3} \theta\), then \(\frac{d^{2} y}{d x^{2}}\) at \(\theta=\frac{\pi}{2}\) is

1 \(\frac{1}{6}\)
2 3
3 6
4 \(\frac{1}{3}\)
MHT CET-2019
Limits, Continuity and Differentiability

80265 If \(y=\tan ^{-1}\left(\frac{1-\cos 3 x}{\sin 3 x}\right)\), then \(\frac{d y}{d x}=\)

1 \(-\frac{3}{2}\)
2 \(\frac{1}{2}\)
3 \(-\frac{1}{2}\)
4 \(\frac{3}{2}\)
MHT CET-2019
Limits, Continuity and Differentiability

80266 If \(y=\log \left|\frac{x+\sqrt{x^{2}+25}}{\sqrt{x^{2}+25}-x}\right|\), then \(\frac{d y}{d x}=\)

1 \(\frac{-1}{\sqrt{\mathrm{x}^{2}+25}}\)
2 \(\frac{-2}{\sqrt{\mathrm{x}^{2}+25}}\)
3 \(\frac{2}{\sqrt{\mathrm{x}^{2}+25}}\)
4 \(\frac{1}{\sqrt{\mathrm{x}^{2}+25}}\)
MHT CET-2019
Limits, Continuity and Differentiability

80267 Derivative of \(\sin ^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)\) with respect to \(\cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)\) is

1 \(\tan t\)
2 0
3 1
4 \(\cot \mathrm{t}\)
MHT CET-2019
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Limits, Continuity and Differentiability

80262 If \(x=\sqrt{\mathbf{a}^{\sin ^{-1} t}}, y=\sqrt{\mathbf{a}^{\cos ^{-1} t}}\), then \(\frac{d y}{d x}=\)

1 \(\frac{-x}{y}\)
2 \(\frac{y}{x}\)
3 \(\frac{-y}{x}\)
4 \(\frac{x}{y}\)
MHT CET-2019
Limits, Continuity and Differentiability

80264 If \(x=\sin \theta, y=\sin ^{3} \theta\), then \(\frac{d^{2} y}{d x^{2}}\) at \(\theta=\frac{\pi}{2}\) is

1 \(\frac{1}{6}\)
2 3
3 6
4 \(\frac{1}{3}\)
MHT CET-2019
Limits, Continuity and Differentiability

80265 If \(y=\tan ^{-1}\left(\frac{1-\cos 3 x}{\sin 3 x}\right)\), then \(\frac{d y}{d x}=\)

1 \(-\frac{3}{2}\)
2 \(\frac{1}{2}\)
3 \(-\frac{1}{2}\)
4 \(\frac{3}{2}\)
MHT CET-2019
Limits, Continuity and Differentiability

80266 If \(y=\log \left|\frac{x+\sqrt{x^{2}+25}}{\sqrt{x^{2}+25}-x}\right|\), then \(\frac{d y}{d x}=\)

1 \(\frac{-1}{\sqrt{\mathrm{x}^{2}+25}}\)
2 \(\frac{-2}{\sqrt{\mathrm{x}^{2}+25}}\)
3 \(\frac{2}{\sqrt{\mathrm{x}^{2}+25}}\)
4 \(\frac{1}{\sqrt{\mathrm{x}^{2}+25}}\)
MHT CET-2019
Limits, Continuity and Differentiability

80267 Derivative of \(\sin ^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)\) with respect to \(\cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)\) is

1 \(\tan t\)
2 0
3 1
4 \(\cot \mathrm{t}\)
MHT CET-2019
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Limits, Continuity and Differentiability

80262 If \(x=\sqrt{\mathbf{a}^{\sin ^{-1} t}}, y=\sqrt{\mathbf{a}^{\cos ^{-1} t}}\), then \(\frac{d y}{d x}=\)

1 \(\frac{-x}{y}\)
2 \(\frac{y}{x}\)
3 \(\frac{-y}{x}\)
4 \(\frac{x}{y}\)
MHT CET-2019
Limits, Continuity and Differentiability

80264 If \(x=\sin \theta, y=\sin ^{3} \theta\), then \(\frac{d^{2} y}{d x^{2}}\) at \(\theta=\frac{\pi}{2}\) is

1 \(\frac{1}{6}\)
2 3
3 6
4 \(\frac{1}{3}\)
MHT CET-2019
Limits, Continuity and Differentiability

80265 If \(y=\tan ^{-1}\left(\frac{1-\cos 3 x}{\sin 3 x}\right)\), then \(\frac{d y}{d x}=\)

1 \(-\frac{3}{2}\)
2 \(\frac{1}{2}\)
3 \(-\frac{1}{2}\)
4 \(\frac{3}{2}\)
MHT CET-2019
Limits, Continuity and Differentiability

80266 If \(y=\log \left|\frac{x+\sqrt{x^{2}+25}}{\sqrt{x^{2}+25}-x}\right|\), then \(\frac{d y}{d x}=\)

1 \(\frac{-1}{\sqrt{\mathrm{x}^{2}+25}}\)
2 \(\frac{-2}{\sqrt{\mathrm{x}^{2}+25}}\)
3 \(\frac{2}{\sqrt{\mathrm{x}^{2}+25}}\)
4 \(\frac{1}{\sqrt{\mathrm{x}^{2}+25}}\)
MHT CET-2019
Limits, Continuity and Differentiability

80267 Derivative of \(\sin ^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)\) with respect to \(\cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)\) is

1 \(\tan t\)
2 0
3 1
4 \(\cot \mathrm{t}\)
MHT CET-2019
NEET DIGITAL LIBRARY
Limits, Continuity and Differentiability

80262 If \(x=\sqrt{\mathbf{a}^{\sin ^{-1} t}}, y=\sqrt{\mathbf{a}^{\cos ^{-1} t}}\), then \(\frac{d y}{d x}=\)

1 \(\frac{-x}{y}\)
2 \(\frac{y}{x}\)
3 \(\frac{-y}{x}\)
4 \(\frac{x}{y}\)
MHT CET-2019
Limits, Continuity and Differentiability

80264 If \(x=\sin \theta, y=\sin ^{3} \theta\), then \(\frac{d^{2} y}{d x^{2}}\) at \(\theta=\frac{\pi}{2}\) is

1 \(\frac{1}{6}\)
2 3
3 6
4 \(\frac{1}{3}\)
MHT CET-2019
Limits, Continuity and Differentiability

80265 If \(y=\tan ^{-1}\left(\frac{1-\cos 3 x}{\sin 3 x}\right)\), then \(\frac{d y}{d x}=\)

1 \(-\frac{3}{2}\)
2 \(\frac{1}{2}\)
3 \(-\frac{1}{2}\)
4 \(\frac{3}{2}\)
MHT CET-2019
Limits, Continuity and Differentiability

80266 If \(y=\log \left|\frac{x+\sqrt{x^{2}+25}}{\sqrt{x^{2}+25}-x}\right|\), then \(\frac{d y}{d x}=\)

1 \(\frac{-1}{\sqrt{\mathrm{x}^{2}+25}}\)
2 \(\frac{-2}{\sqrt{\mathrm{x}^{2}+25}}\)
3 \(\frac{2}{\sqrt{\mathrm{x}^{2}+25}}\)
4 \(\frac{1}{\sqrt{\mathrm{x}^{2}+25}}\)
MHT CET-2019
Limits, Continuity and Differentiability

80267 Derivative of \(\sin ^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)\) with respect to \(\cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)\) is

1 \(\tan t\)
2 0
3 1
4 \(\cot \mathrm{t}\)
MHT CET-2019
Join With Us for More Updates
Limits, Continuity and Differentiability

80262 If \(x=\sqrt{\mathbf{a}^{\sin ^{-1} t}}, y=\sqrt{\mathbf{a}^{\cos ^{-1} t}}\), then \(\frac{d y}{d x}=\)

1 \(\frac{-x}{y}\)
2 \(\frac{y}{x}\)
3 \(\frac{-y}{x}\)
4 \(\frac{x}{y}\)
MHT CET-2019
Limits, Continuity and Differentiability

80264 If \(x=\sin \theta, y=\sin ^{3} \theta\), then \(\frac{d^{2} y}{d x^{2}}\) at \(\theta=\frac{\pi}{2}\) is

1 \(\frac{1}{6}\)
2 3
3 6
4 \(\frac{1}{3}\)
MHT CET-2019
Limits, Continuity and Differentiability

80265 If \(y=\tan ^{-1}\left(\frac{1-\cos 3 x}{\sin 3 x}\right)\), then \(\frac{d y}{d x}=\)

1 \(-\frac{3}{2}\)
2 \(\frac{1}{2}\)
3 \(-\frac{1}{2}\)
4 \(\frac{3}{2}\)
MHT CET-2019
Limits, Continuity and Differentiability

80266 If \(y=\log \left|\frac{x+\sqrt{x^{2}+25}}{\sqrt{x^{2}+25}-x}\right|\), then \(\frac{d y}{d x}=\)

1 \(\frac{-1}{\sqrt{\mathrm{x}^{2}+25}}\)
2 \(\frac{-2}{\sqrt{\mathrm{x}^{2}+25}}\)
3 \(\frac{2}{\sqrt{\mathrm{x}^{2}+25}}\)
4 \(\frac{1}{\sqrt{\mathrm{x}^{2}+25}}\)
MHT CET-2019
Limits, Continuity and Differentiability

80267 Derivative of \(\sin ^{-1}\left(\frac{t}{\sqrt{1+t^{2}}}\right)\) with respect to \(\cos ^{-1}\left(\frac{1}{\sqrt{1+t^{2}}}\right)\) is

1 \(\tan t\)
2 0
3 1
4 \(\cot \mathrm{t}\)
MHT CET-2019