Question

If \[\sin y + {e^{ - x\cos y}} = e\]. Then what is the value of \[\dfrac{{dy}}{{dx}}\] at \[\left( {1,\pi } \right)\]?
A. \[e\]
B. \[\sin y\]
C. \[\cos y\]
D. \[\sin y \cos y\]

Answer 1

Hint In the given question, one trigonometric equation is given. We will differentiate the given equations with respect to \[x\]. Then simplify the differential equation. By substituting \[x = 1\] and \[y = \pi \] in the differential equation, we will get the value of \[\dfrac{{dy}}{{dx}}\] at \[\left( {1,\pi } \right)\].

Formula used
\[\dfrac{d}{{dx}}\left( {\sin y} \right) = \cos y\dfrac{{dy}}{{dx}}\]
\[\dfrac{d}{{dx}}\left( {\cos y} \right) = - \sin y\dfrac{{dy}}{{dx}}\]
\[\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\]
\[\dfrac{d}{{dx}}\left( c \right) = 0\], where \[c\] is a constant.
Product rule formula: \[\dfrac{d}{{dx}}\left( {uv} \right) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\]

Complete step by step solution:
The given trigonometric equation is \[\sin y + {e^{ - x\cos y}} = e\].
Let’s differentiate the given equation with respect to \[x\].
\[\dfrac{d}{{dx}}\sin y + \dfrac{d}{{dx}}{e^{ - x\cos y}} = \dfrac{d}{{dx}}e\]
\[ \Rightarrow \]\[\cos y\dfrac{{dy}}{{dx}} + {e^{ - x\cos y}}\dfrac{d}{{dx}}\left( { - x\cos y} \right) = 0\]
Apply product rule formula for the exponential term.
\[\cos y\dfrac{{dy}}{{dx}} + {e^{ - x\cos y}}\left( { - \cos y + x\sin y\dfrac{{dy}}{{dx}}} \right) = 0\] [Since \[e\] is a constant. So, \[\dfrac{d}{{dx}}\left( e \right) = 0\]]
Simplify the above equation.
\[\left( {\cos y + {e^{ - x\cos y}}x\sin y} \right)\dfrac{{dy}}{{dx}} - {e^{ - x\cos y}}\cos y = 0\]
\[ \Rightarrow \]\[\left( {\cos y + {e^{ - x\cos y}}x\sin y} \right)\dfrac{{dy}}{{dx}} = {e^{ - x\cos y}}\cos y\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{{{e^{ - x\cos y}}\cos y}}{{\left( {\cos y + {e^{ - x\cos y}}x\sin y} \right)}}\]

Now substitute \[x = 1\] and \[y = \pi \] in the above equation.
\[\dfrac{{dy}}{{dx}} = \dfrac{{{e^{ - \left( 1 \right)\cos\left( \pi \right)}}\cos\left( \pi \right)}}{{\left( {\cos\left( \pi \right) + {e^{ - \left( 1 \right)\cos\left( \pi \right)}}\left( 1 \right)\sin\left( \pi \right)} \right)}}\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{{{e^{ - \left( 1 \right)\left( { - 1} \right)}}\left( { - 1} \right)}}{{\left( {\left( { - 1} \right) + {e^{ - \left( 1 \right)\left( { - 1} \right)}}\left( 1 \right)\left( 0 \right)} \right)}}\] [Since \[\cos\pi = - 1\] and \[\sin\pi = 0\]]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = \dfrac{{ - e}}{{ - 1}}\]
\[ \Rightarrow \]\[\dfrac{{dy}}{{dx}} = e\]

Hence the correct option is A.

Note: Students are often confused with the formulas of \[\dfrac{d}{{dx}}\left( {\sin y} \right)\] and \[\dfrac{d}{{dx}}\left( {\cos y} \right)\]. Here \[y\] is another variable, so we have to also calculate the derivative of the variable \[y\]. The derivative of \[\dfrac{d}{{dx}}\left( {\sin y} \right)\] is \[\cos y\dfrac{{dy}}{{dx}}\] and \[\dfrac{d}{{dx}}\left( {\cos y} \right)\] is \[ - \sin y\dfrac{{dy}}{{dx}}\].