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MHTCET 10 May 2024 Shift 1 Paper

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Question : 136 of 150

Marks: +1, -0

If

y

is a function of

x

\log (x+y)=2 x y

, then the value of

y^{\prime}(0)

is

1
-1
2
0

Solution:

Concept:

Use implicit differentiation to find

y'

, then substitute

x=0

after finding the corresponding

y

.

Explanation:

Given:

\log(x+y)=2xy

.

At

x=0

, we get

\log(y)=0

, so

y=1

.

Differentiate both sides with respect to

x

:

\frac{1}{x+y}\left(1+y'\right)=2\left(xy'+y\right)

Multiply both sides by

(x+y)

:

1+y'=2x(x+y)y'+2y(x+y)

Rearrange to isolate

y'

:

y'=\frac{1-2y(x+y)}{2x(x+y)-1}

Now substitute

x=0

and

y=1

:

y'(0)=\frac{1-2(1)(0+1)}{2(0)(0+1)-1}

y'(0)=\frac{1-2}{-1}=\frac{-1}{-1}=1

Answer:

y'(0)=1

Correct option: A. 1

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