Differentiation Question 216
Question: If $ A=\frac{2^{x}\cot x}{\sqrt{x}}, $ then $ \frac{dA}{dx}= $
Options:
A) $ \frac{{2^{x-1}}{ -2xcose{c^{2}}x+\cot x.\log ( \frac{4^{x}}{e} ) }}{{x^{3/2}}} $
B) $ \frac{{2^{x-1}}{ -2x\cos e{c^{2}}x+\cot x.\log ( \frac{4^{x}}{e} ) }}{x} $
C) $ \frac{2x{ -2xcose{c^{2}}x+\cot x.\log ( \frac{4^{x}}{e} ) }}{{x^{\text{3/2}}}} $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ \frac{dA}{dx}=\frac{\sqrt{x}{2^{x}{\log _{e}}2\cot x-2^{x}cose{c^{2}}x}-2^{x}\cot x\frac{1}{2\sqrt{x}}}{x} $
$ =\frac{{2^{x-1}}{ -2xcose{c^{2}}x+\cot x.\log ( \frac{4^{x}}{e} ) }}{{x^{3/2}}} $ .
BETA
