Question

If $y{\text{ is a function of }}x$ and $\log \left( {x + y} \right) = 2xy$ , then the value of $y'\left( 0 \right)$ is :
$\left( 1 \right)1$
$\left( 2 \right) - 1$
$\left( 3 \right)2$
$\left( 4 \right)0$

Answer 1

Hint: To solve this question, we should be familiar with standard differentiation properties and identities of the logarithmic function. The derivative of $\dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}$ . There can be multiple cases like this, let say the argument of the logarithmic function is a polynomial. For example: $\dfrac{d}{{dx}}\log \left( {2{x^3} + 1} \right)$ then the differentiation will be $\dfrac{{6{x^2}}}{{x\log \left( {2{x^3} + 1} \right)}}$ . So, by the proper knowledge of properties of logarithms and the application of chain rule, these types of problems become easier to solve. Although there is one strict condition for usage of logarithmic formula i.e. the individual functions must be
positive since log function is defined for only positive values.

Complete step by step answer:
According to the question;
$ \Rightarrow \log \left( {x + y} \right) = 2xy$
Differentiate both sides w.r.t. $x$ , we get;
$ \Rightarrow \dfrac{1}{{\left( {x + y} \right)}}\left( {1 + \dfrac{{dy}}{{dx}}} \right) = 2\left( {1 \times y + x \times \dfrac{{dy}}{{dx}}} \right){\text{ }}......\left( 1 \right)$
[we know that $ \Rightarrow x + y = 1 + \dfrac{{dy}}{{dx}}$
$ \Rightarrow 2xy = 2\left( {1 \times y + x \times \dfrac{{dy}}{{dx}}} \right)$
we know that $\dfrac{d}{{dx}}\left( {uv} \right) = v\dfrac{{du}}{{dx}} + u\dfrac{{dv}}{{dx}}$
$\left( {\because \dfrac{d}{{dx}}\left( {\log x} \right) = \dfrac{1}{x}} \right)$
Let $\dfrac{{dy}}{{dx}} = y'\left( x \right)$]
Further simplifying equation $\left( 1 \right)$ , we get;
$ \Rightarrow \dfrac{1}{{\left( {x + y} \right)}} + y'\left( x \right)\left( {\dfrac{1}{{x + y}}} \right){\text{ = 2}}y{\text{ + 2}}xy'\left( x \right)$
Taking all the $y'\left( x \right)$ terms to the L.H.S. , we get;
$ \Rightarrow y'\left( x \right)\left( {\dfrac{1}{{x + y}}} \right) - y'\left( x \right){\text{2}}x = 2y - \dfrac{1}{{x + y}}{\text{ }}$
Taking $y'\left( x \right)$ common in the L.H.S. ;
$ \Rightarrow y'\left( x \right)\left[ {\dfrac{1}{{x + y}} - 2x} \right] = 2y - \dfrac{1}{{x + y}}{\text{ }}......\left( 2 \right)$
$\because \log \left( {x + y} \right) = 2xy{\text{ }}\left( {{\text{given in the question}}} \right)$
Put $x = 0$ , in the above equation, we get;
$ \Rightarrow \log \left( {0 + y} \right) = 0$
$ \Rightarrow \log \left( y \right) = 0$
Therefore, we can clearly say that the value of $y$ will be,
$ \Rightarrow y = 1{\text{ }}\left( {\because \log 1 = 0} \right)$
Now, we have both the values i.e.${\text{when }}x = 0,{\text{ then }}y = 1$
Put the values of $x$ and $y$ in equation $\left( 2 \right)$ , we get;
$ \Rightarrow y'\left( 0 \right)\left[ {\dfrac{1}{{0 + 1}} - 2 \times 0} \right] = 2 \times 1 - \dfrac{1}{{0 + 1}}$
$ \Rightarrow y'\left( 0 \right)\left[ 1 \right] = 2 - 1$
And finally we get the value of $y'\left( 0 \right)$ as;
$ \Rightarrow y'\left( 0 \right) = 1$

So, the correct answer is “Option 1”.

Note: In the given question they have asked to find the derivative and then replace the argument value as $0$ , which basically means differential value of a function at a point $x = 0$ . We know that the differentiation of a function represents it’s rate of change with respect to some variable, notice here that we have got the value of ${\left. {\dfrac{{dy}}{{dx}}} \right|_{x = 0}}{\text{ is 1}}$ , which practically means that around the point $x = 0$ , our function is almost constant. This case happens if there exists a derivative for a function, they always change gradually from one point to another point, that is why whenever we calculate their derivative value at $x = 0$ , they show almost constant value or no change.