Question

If $\sin y + {e^{ - x\cos y}} = e$, then $\dfrac{{dy}}{{dx}}$ at $\left( {1,\pi } \right)$ is
(A) $e$
(B) $\sin y$
(C) $\cos y$
(D) $\sin y\cos y$

Answer 1

Hint: In the given problem, we are required to differentiate the implicit function $\sin y + {e^{ - x\cos y}} = e$ with respect to x and find the value of derivative for point lying on the curve $\left( {1,\pi } \right)$. Since, $\sin y + {e^{ - x\cos y}} = e$ is an implicit function, so we will have to differentiate the function with the implicit method of differentiation. So, differentiation of $\sin y + {e^{ - x\cos y}} = e$ with respect to x will be done layer by layer using the chain rule of differentiation as in the given function, we cannot isolate the variables x and y. After differentiating, we will put the given values of x and y to get to the final answer.

Complete step-by-step answer:
So, we have, $\sin y + {e^{ - x\cos y}} = e$
Differentiating both sides of the equation with respect to x, we get,
$\dfrac{d}{{dx}}\left( {\sin y} \right) + \dfrac{d}{{dx}}\left( {{e^{ - x\cos y}}} \right) = \dfrac{d}{{dx}}\left( e \right)$
We know that derivatives of constant terms is equal to zero. So, we get,
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sin y} \right) + \dfrac{d}{{dx}}\left( {{e^{ - x\cos y}}} \right) = 0$
Now, we use the chain rule of differentiation \[\dfrac{d}{{dx}}\left[ {f\left( {g\left( x \right)} \right)} \right] = f'\left( {g\left( x \right)} \right) \times g'\left( x \right)\], we get,
$ \Rightarrow \cos y \times \dfrac{{dy}}{{dx}} + \dfrac{d}{{dx}}\left( {{e^{ - x\cos y}}} \right) = 0$
Now, we know that derivative of exponential function ${e^x}$ is ${e^x}$. So, we get,
$ \Rightarrow \cos y \times \dfrac{{dy}}{{dx}} + {e^{ - x\cos y}} \times \dfrac{{d\left( { - x\cos y} \right)}}{{dx}} = 0$
Using product rule of differentiation, we get,
\[ \Rightarrow \cos y \times \dfrac{{dy}}{{dx}} + {e^{ - x\cos y}} \times \left[ { - x \times \dfrac{{d\left( {\cos y} \right)}}{{dx}} + \left( {\cos y} \right) \times \dfrac{{d\left( { - x} \right)}}{{dx}}} \right] = 0\]
Simplifying the expression, we get,
\[ \Rightarrow \cos y \times \dfrac{{dy}}{{dx}} + {e^{ - x\cos y}} \times \left[ {x\sin y \times \dfrac{{dy}}{{dx}} - \cos y} \right] = 0\]
Now, opening the brackets,
\[ \Rightarrow \cos y \times \dfrac{{dy}}{{dx}} + \left( {{e^{ - x\cos y}}} \right)x\sin y \times \dfrac{{dy}}{{dx}} - \cos y\left( {{e^{ - x\cos y}}} \right) = 0\]
Taking $\dfrac{{dy}}{{dx}}$ common from terms, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}}\left[ {\cos y + \left( {{e^{ - x\cos y}}} \right)x\sin y} \right] - \cos y\left( {{e^{ - x\cos y}}} \right) = 0\]
Isolating the term $\dfrac{{dy}}{{dx}}$,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos y\left( {{e^{ - x\cos y}}} \right)}}{{\cos y + \left( {{e^{ - x\cos y}}} \right)x\sin y}}\]
Putting the values $x = 1$ and $y = \pi $, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\cos \pi \left( {{e^{ - \left( 1 \right)\cos \pi }}} \right)}}{{\cos \pi + \left( {{e^{ - \left( 1 \right)\cos \pi }}} \right)\left( 1 \right)\sin \pi }}\]
Substituting the values of sine and cosine for standard angles, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{ - e}}{{ - 1}}\]
Cancelling common factors in numerator and denominator, we get,
\[ \Rightarrow \dfrac{{dy}}{{dx}} = e\]
So, we get the value of \[\dfrac{{dy}}{{dx}}\] as $e$.
Therefore, option (A) is the correct answer.
So, the correct answer is “Option A”.

Note: Implicit functions are those functions that involve two variables and the two variables are not separable and cannot be isolated from each other. Hence, we have to follow a certain method to differentiate such functions and also apply the chain rule of differentiation. We must remember the simple derivatives of basic functions to solve such problems. One should use the simplification rules in order to simplify the expression. We should make sure that the calculations are correct to get to the required answer.