Match List-I with List-II
|
List-I Function |
List-II Derivative |
|
(A) $y = e^{3\log_e3}$ |
(I) $\frac{dy}{dx}=\frac{1}{2y-1}$ |
|
(B) $y=\sqrt{x+y},x+y> 0$ and $y≠\frac{1}{2}$ |
(II) $\frac{dy}{dx}=10^x\log_e10$ |
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(C) $y = \log_{10} x, x>0$ |
(III) $\frac{dy}{dx} =0$ |
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(D) $y = 10^x$ |
(IV) $\frac{dy}{dx}=\frac{1}{x\log_e10}$ |
Choose the correct answer from the options given below:
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → (A)-(III), (B)-(I), (C)-(IV), (D)-(II)
|
List-I Function |
List-II Derivative |
|
(A) $y = e^{3\log_e3}$ |
(III) $\frac{dy}{dx} =0$ |
|
(B) $y=\sqrt{x+y},x+y> 0$ and $y≠\frac{1}{2}$ |
(I) $\frac{dy}{dx}=\frac{1}{2y-1}$ |
|
(C) $y = \log_{10} x, x>0$ |
(IV) $\frac{dy}{dx}=\frac{1}{x\log_e10}$ |
|
(D) $y = 10^x$ |
(II) $\frac{dy}{dx}=10^x\log_e10$ |
$(A)\; y= e^{3\log_e3} = 3^3=27$
Differentiate:
$\frac{dy}{dx} =0$
Hence derivative is $0$ → matches (III).
$(B)\; y=\sqrt{x+y},\ x+y>0,\ y\neq\frac12$
Square both sides: $y^{2}=x+y$
Differentiate: $2y\frac{dy}{dx}=1+\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{1}{2y-1}$
Matches (I).
$(C)\; y=\log_{10}x,\ x>0$
$\frac{dy}{dx}=\frac{1}{x\ln 10}$
Matches (IV).
$(D)\; y=10^{x}$
$\frac{dy}{dx}=10^{x}\ln 10$
Matches (II).
Thus, the correct answer is: $(A\!-\!III),\ (B\!-\!I),\ (C\!-\!IV),\ (D\!-\!II)$.