Question

Differentiate with respect to x.
\[y = \dfrac{1}{{\sin x}} + {e^x} + 2\]

Answer 1

Hint: For this type of problem having different functions. We first write them as in addition or subtraction if possible and then apply respective formulas of differentiation on each to get the required solution.
Quotient rule of differentiation: $ \dfrac{d}{{dx}}\left( {\dfrac{A}{B}} \right) = \dfrac{{B.\dfrac{d}{{dx}}(A) - A\dfrac{d}{{dx}}(B)}}{{{B^2}}} $ , $ \dfrac{d}{{dx}}(\sin x) = \cos x,\,\,\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\,\,and\,\,\dfrac{d}{{dx}}\left( {\operatorname{co} ns\tan t} \right) = 0 $

Complete step by step solution:
 \[y = \dfrac{1}{{\sin x}} + {e^x} + 2\]
Differentiating above with respect to x
\[\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{1}{{\sin x}} + {e^x} + 2} \right)\]
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {\dfrac{1}{{\sin x}}} \right) + \dfrac{d}{{dx}}\left( {{e^x}} \right) + \dfrac{d}{{dx}}\left( 2 \right) $
Applying, quotient rule on first term of right hand side of the above equation to find differentiation.
 $ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\sin x\dfrac{d}{{dx}}(1) - 1\dfrac{d}{{dx}}(\sin x)}}{{{{\sin }^2}x}} + \dfrac{d}{{dx}}\left( {{e^x}} \right) + \dfrac{d}{{dx}}(2) $
 $
  \dfrac{{dy}}{{dx}} = \dfrac{{\sin x(0) - \cos x}}{{{{\sin }^2}x}} + {e^x} + 0 \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{{\cos x}}{{{{\sin }^2}x}} + {e^x} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \dfrac{1}{{\sin x}}.\dfrac{{\cos x}}{{\sin x}} + {e^x} \\
   \Rightarrow \dfrac{{dy}}{{dx}} = - \cos ecx.\cot x + {e^x} \\
  $
Hence, from above we see that differentiation of $ y = \dfrac{1}{{\sin x}} + {e^x} + 2 $ w.r.t. x is $ - \cos ecx.\cot x + {e^x} $ .
So, the correct answer is “$ - \cos ecx.\cot x + {e^x} $”.

Note: For a given problem we can also find its answer in another way. In this we first write $ \dfrac{1}{{\sin x}} $ as $ \cos ecx $ and then the right hand side of the given problem becomes the sum of three independent functions. After differentiating them individually we can find the derivative of the given problem.