Question

If $y = y(x)$ and it follows the relation ${e^{x{y^2}}} + y\cos ({x^2}) = 5$ then $y'(0)$ is equal to
A. 4
B. -16
C. -4
D. 16

Answer 1

Hint: $y$ is a function of $x$. First find the value of $y(0)$ by substituting $x = 0$ in the equation. Then differentiate the equation with respect to $x$ and use the product rule for differentiating products of different terms in the equation. Substitute $x = 0$ and value of $y(0)$in the final result.

Complete step-by-step answer:
Let us consider the equation:
${e^{x{y^2}}} + y\cos ({x^2}) = 5$. We have $x$ and $y$ terms in the equation. Since, y is a function of $x$, we will find its value for $x = 0$. After substituting $x = 0$ in the above equation, we get,
$
   \Rightarrow {e^{0 \times {{[y(0)]}^2}}} + y(0)\cos (0) = 5 \\
   \Rightarrow 1 + y(0) = 5 \\
   \Rightarrow y(0) = 4 \\
$ (1)
Now we have a value of $y$ for $x = 0$. When we differentiate the above equation with respect to $x$, we will get another equation but with additional $y' = \dfrac{{dy}}{{dx}}$ terms.
Differentiating given equation with respect to $x$, we get,
$
   \Rightarrow \dfrac{{d[{e^{x{y^2}}} + y\cos ({x^2})]}}{{dx}} = \dfrac{{d(5)}}{{dx}} \\
   \Rightarrow \dfrac{{d[{e^{x{y^2}}}]}}{{dx}} + \dfrac{{d[y\cos ({x^2})]}}{{dx}} = 0 \\
$
Here, we know that derivative of a constant, here 5, is always zero. Another rule we used is that the derivative of the sum of terms is equal to the sum of derivatives of the terms. To solve it further we will use here product rule which is as follows: $\dfrac{{d[u(x)v(x)]}}{{dx}} = v(x)\dfrac{{du(x)}}{{dx}} + u(x)\dfrac{{dv(x)}}{{dx}}$.
For the first term, we have an exponential term which is in the form: $\dfrac{{d[{e^{f(x)}}]}}{{dx}} = {e^{f(x)}}\dfrac{{df(x)}}{{dx}}$
Solving the problem further, we get,
$ \Rightarrow \underbrace {{e^{x{y^2}}}\dfrac{{d[x{y^2}]}}{{dx}}}_{{1^{st}}term} + \underbrace {\cos ({x^2})\dfrac{{dy}}{{dx}} + y\dfrac{{d\cos ({x^2})}}{{dx}}}_{{2^{nd}}term} = 0$
The last term above is in the form, $\dfrac{{dg[f(x)]}}{{dx}} = \dfrac{{dg[f(x)]}}{{df(x)}} \times \dfrac{{df(x)}}{{dx}}$
$
   \Rightarrow {e^{x{y^2}}}\left[ {{y^2}\dfrac{{dx}}{{dx}} + x\dfrac{{d{y^2}}}{{dx}}} \right] + y'\cos ({x^2}) - y\sin ({x^2})\dfrac{{d{x^2}}}{{dx}} = 0 \\
   \Rightarrow {e^{x{y^2}}}\left[ {{y^2} + 2xyy'} \right] + y'\cos ({x^2}) - 2xy\sin ({x^2}) = 0 \\
$
From above result and (1), we will substitute $x = 0$ and thus we get,
$
   \Rightarrow {e^{0 \times {4^2}}}[{4^2} + (2 \times 0 \times 4 \times y')] + y'\cos (0) - [2 \times 0 \times 4 \times \sin (0)] = 0 \\
   \Rightarrow 16 + y' = 0 \\
   \Rightarrow y' = - 16 \\
 $
After comparing the above answer with the given choices, option B is the correct answer.

Note: Please make sure you start practicing with writing all the steps correctly as for the above step in exponential form or the other of the functions inside a function. There are very high chances of missing steps in between and ending up with completely different solutions. Also make sure the powers and symbols for differentiation are well separated.