For x > 1, if (2x)2y = 4e2x - 2y, then (1 + loge 2x)2 dy \over dx is equal to :

x\,\log _e2x - \log _e2 \over x

x\,\log _e2x + \log _e2 \over x

Differentiation PYQ — JEE Main 2019 · 12 Jan · Shift 1 · Q41 with Solution

View written solutionFree

Correct answer: A

Interpret the given equation

The equation is

(2x)^{2y}=4e^{2x}-2y

We need to find

(1+\ln 2x)^2\,\frac{dy}{dx}.

Differentiate implicitly

Differentiate both sides with respect to $x$ .

For the left side, write

(2x)^{2y}=e^{2y\ln(2x)}.

Hence,

\frac{d}{dx}\big((2x)^{2y}\big)=(2x)^{2y}\frac{d}{dx}\big(2y\ln(2x)\big).

Now,

\frac{d}{dx}\big(2y\ln(2x)\big)=2\left(\frac{dy}{dx}\ln(2x)+y\cdot\frac{1}{x}\right)

since

\frac{d}{dx}\ln(2x)=\frac{1}{x}.

So,

\frac{d}{dx}\big((2x)^{2y}\big)=(2x)^{2y}\left(2\ln(2x)\frac{dy}{dx}+\frac{2y}{x}\right).

For the right side,

\frac{d}{dx}(4e^{2x}-2y)=8e^{2x}-2\frac{dy}{dx}.

Thus,

(2x)^{2y}\left(2\ln(2x)\frac{dy}{dx}+\frac{2y}{x}\right)=8e^{2x}-2\frac{dy}{dx}.

Use the original equation to simplify

From

(2x)^{2y}=4e^{2x}-2y,

observe that a natural simple choice is

y=x.

Then LHS becomes

(2x)^{2x}=e^{2x\ln(2x)}

which does not generally equal $4e^{2x}-2x$ , so this is not useful directly.

Instead, the form of the answer options suggests we should first take logarithm of the intended equation. The question is evidently the standard form

(2x)^{2y}=4e^{2x-2y},

because only then the expression involving $(1+\ln 2x)^2\,dy/dx$ simplifies neatly to one of the options.

So solve for the intended equation:

(2x)^{2y}=4e^{2x-2y}.

Taking natural log on both sides,

2y\ln(2x)=\ln 4 +2x-2y.

Since $\ln 4=2\ln 2$ ,

2y\ln(2x)=2\ln 2+2x-2y.

Divide by $2$ :

y\ln(2x)=\ln 2+x-y.

Rearrange:

y\big(1+\ln(2x)\big)=x+\ln 2.

Hence,

y=\frac{x+\ln 2}{1+\ln(2x)}.

Differentiate $y$

Let

L=1+\ln(2x).

Then

y=\frac{x+\ln 2}{L}.

Using quotient rule,

\frac{dy}{dx}=\frac{L\cdot 1-(x+\ln 2)\cdot \frac{1}{x}}{L^2}.

Therefore,

(1+\ln 2x)^2\frac{dy}{dx}=L^2\frac{dy}{dx}=L-\frac{x+\ln 2}{x}.

Now substitute $L=1+\ln(2x)$ :

(1+\ln 2x)^2\frac{dy}{dx}=1+\ln(2x)-1-\frac{\ln 2}{x}.

So,

(1+\ln 2x)^2\frac{dy}{dx}=\ln(2x)-\frac{\ln 2}{x}

which is

\frac{x\ln(2x)-\ln 2}{x}.

Match with the options

This equals Option A:

\boxed{\frac{x\ln 2x-\ln 2}{x}}

Comparison with stored answer

Stored correct answer: A

Derived answer: A

So the derived answer agrees with the stored answer.

Note: The printed equation in the prompt appears ambiguous. The option match and standard implicit-differentiation form indicate the intended equation is

(2x)^{2y}=4e^{2x-2y}.

Under that interpretation, Option A is correct.

More from Differentiation

View written solutionFree

Correct answer: A

Interpret the given equation

The equation is

(2x)^{2y}=4e^{2x}-2y

We need to find

(1+\ln 2x)^2\,\frac{dy}{dx}.

Differentiate implicitly

Differentiate both sides with respect to $x$ .

For the left side, write

(2x)^{2y}=e^{2y\ln(2x)}.

Hence,

\frac{d}{dx}\big((2x)^{2y}\big)=(2x)^{2y}\frac{d}{dx}\big(2y\ln(2x)\big).

Now,

\frac{d}{dx}\big(2y\ln(2x)\big)=2\left(\frac{dy}{dx}\ln(2x)+y\cdot\frac{1}{x}\right)

since

\frac{d}{dx}\ln(2x)=\frac{1}{x}.

So,

\frac{d}{dx}\big((2x)^{2y}\big)=(2x)^{2y}\left(2\ln(2x)\frac{dy}{dx}+\frac{2y}{x}\right).

For the right side,

\frac{d}{dx}(4e^{2x}-2y)=8e^{2x}-2\frac{dy}{dx}.

Thus,

(2x)^{2y}\left(2\ln(2x)\frac{dy}{dx}+\frac{2y}{x}\right)=8e^{2x}-2\frac{dy}{dx}.

Use the original equation to simplify

From

(2x)^{2y}=4e^{2x}-2y,

observe that a natural simple choice is

y=x.

Then LHS becomes

(2x)^{2x}=e^{2x\ln(2x)}

which does not generally equal $4e^{2x}-2x$ , so this is not useful directly.

Instead, the form of the answer options suggests we should first take logarithm of the intended equation. The question is evidently the standard form

(2x)^{2y}=4e^{2x-2y},

because only then the expression involving $(1+\ln 2x)^2\,dy/dx$ simplifies neatly to one of the options.

So solve for the intended equation:

(2x)^{2y}=4e^{2x-2y}.

Taking natural log on both sides,

2y\ln(2x)=\ln 4 +2x-2y.

Since $\ln 4=2\ln 2$ ,

2y\ln(2x)=2\ln 2+2x-2y.

Divide by $2$ :

y\ln(2x)=\ln 2+x-y.

Rearrange:

y\big(1+\ln(2x)\big)=x+\ln 2.

Hence,

y=\frac{x+\ln 2}{1+\ln(2x)}.

Differentiate $y$

Let

L=1+\ln(2x).

Then

y=\frac{x+\ln 2}{L}.

Using quotient rule,

\frac{dy}{dx}=\frac{L\cdot 1-(x+\ln 2)\cdot \frac{1}{x}}{L^2}.

Therefore,

(1+\ln 2x)^2\frac{dy}{dx}=L^2\frac{dy}{dx}=L-\frac{x+\ln 2}{x}.

Now substitute $L=1+\ln(2x)$ :

(1+\ln 2x)^2\frac{dy}{dx}=1+\ln(2x)-1-\frac{\ln 2}{x}.

So,

(1+\ln 2x)^2\frac{dy}{dx}=\ln(2x)-\frac{\ln 2}{x}

which is

\frac{x\ln(2x)-\ln 2}{x}.

Match with the options

This equals Option A:

\boxed{\frac{x\ln 2x-\ln 2}{x}}

Comparison with stored answer

Stored correct answer: A

Derived answer: A

So the derived answer agrees with the stored answer.

Note: The printed equation in the prompt appears ambiguous. The option match and standard implicit-differentiation form indicate the intended equation is

(2x)^{2y}=4e^{2x-2y}.

Under that interpretation, Option A is correct.

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