1
MHT CET 2021 21th September Morning Shift
+2
-0

If $$e^{-y} \cdot y=x$$, then $$\frac{d y}{d x}$$ is

A
$$\frac{y}{1-y}$$
B
$$\frac{1}{x y(1-y)}$$
C
$$\frac{1}{x(1-y)}$$
D
$$\frac{y}{x(1-y)}$$
2
MHT CET 2021 21th September Morning Shift
+2
-0

If $$y=\operatorname{cosec}^{-1}\left[\frac{\sqrt{x}+1}{\sqrt{x}-1}\right]+\cos ^{-1}\left[\frac{\sqrt{x}-1}{\sqrt{x}+1}\right]$$, then $$\frac{d y}{d x}=$$

A
0
B
1
C
$$\frac{2}{\sqrt{x}+1}$$
D
$$\frac{1}{2(\sqrt{x}-1)}$$
3
MHT CET 2021 21th September Morning Shift
+2
-0

The derivative of $$(\log x)^x$$ with respect to $$\log x$$ is

A
$$(\log x)^x\left[\frac{1}{\log x} \log (\log x)\right]$$
B
$$(\log x)^x\left[\log x+\frac{1}{\log (\log x)}\right]$$
C
$$x(\log x)^x\left[\frac{1}{\log x}+\log (\log x)\right]$$
D
$$x(\log x)^x\left[\log x+\frac{1}{\log (\log x)}\right]$$
4
MHT CET 2021 20th September Evening Shift
+2
-0

$$y=\sqrt{e^{\sqrt{x}}}$$, then $$\frac{d y}{d x}=$$

A
$$\frac{e^{\sqrt{x}}}{4 \sqrt{x}}$$
B
$$\frac{\mathrm{e}^{\sqrt{x}}}{4 x}$$
C
$$\frac{e^{\frac{\sqrt{x}}{2}}}{4 \sqrt{x}}$$
D
$$\frac{e^{\sqrt{x}}}{2 \sqrt{x}}$$
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