1
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
$$x(\log x)^x\left[\log x+\frac{1}{\log (\log x)}\right]$$
2
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
$$\frac{e^{\sqrt{x}}}{2 \sqrt{x}}$$
3
MHT CET 2021 20th September Evening Shift
+2
-0

If $$y=\sin ^{-1}\left[\cos \sqrt{\frac{1+x}{2}}\right]+x^x$$, then $$\frac{d y}{d x}$$ at $$x=1$$ is

A
$$\frac{5}{4}$$
B
$$\frac{-1}{4}$$
C
$$\frac{3}{4}$$
D
$$\frac{-5}{4}$$
4
MHT CET 2021 20th September Evening Shift
+2
-0

If $$x=a(t+\sin t), y=a(1-\cos t)$$, then $$\frac{d y}{d x}=$$

A
$$\tan \frac{t}{2}$$
B
$$-\frac{1}{2} \tan \mathrm{t}$$
C
$$\frac{1}{2} \tan \mathrm{t}$$
D
$$-\tan \frac{t}{2}$$
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