Question

Differentiate the following with respect to \[x\]:
\[{{\csc }^{-1}}\left( {{e}^{-x}} \right)\]

Answer 1

Hint: In the given question, we are given a trigonometric function which we have to differentiate with respect to x. We can see that given expression is a composite function, so we will use the chain rule. We will first differentiate the inverse cosecant function and then we will differentiate the exponential term with the negative power of \[x\]. Hence, we will have the derivative of the given expression.

Complete step by step solution:
According to the given question, we are given an expression with an inverse trigonometric function which we have to differentiate with respect to \[x\].
The expression we have is,
\[{{\csc }^{-1}}\left( {{e}^{-x}} \right)\]
Let \[y={{\csc }^{-1}}\left( {{e}^{-x}} \right)\]----(1)
We see that the equation (1) has a composite function, so we will use the chain rule to differentiate the given expression.
We know that the differentiation of the inverse cosecant function is \[\dfrac{d}{dx}\left( {{\csc }^{-1}}x \right)=\dfrac{-1}{x\sqrt{{{x}^{2}}-1}}\].
So, differentiating the equation (1), we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{d}{dx}\left( {{\csc }^{-1}}\left( {{e}^{-x}} \right) \right)\]
So firstly, we will be differentiating the cosecant function and then we will differentiate the exponential function, we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{\left( {{e}^{-x}} \right)\sqrt{{{\left( {{e}^{-x}} \right)}^{2}}-1}}\dfrac{d}{dx}\left( {{e}^{-x}} \right)\]
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{\left( {{e}^{-x}} \right)\sqrt{{{\left( {{e}^{-x}} \right)}^{2}}-1}}\left( {{e}^{-x}} \right)\dfrac{d}{dx}\left( -x \right)\]
Cancelling out the common terms, we have,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{-1}{\sqrt{{{\left( {{e}^{-x}} \right)}^{2}}-1}}\left( -1 \right)\]
The two negative signs will give the expression a positive sign, we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{{{\left( {{e}^{-x}} \right)}^{2}}-1}}\]
We will now square the term and we will get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{{{e}^{-}}^{2x}-1}}\]
We will simplify the terms further, we get the expression as,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{\dfrac{1}{{{e}^{2x}}}-1}}\]
Taking the LCM of the terms in the denominator, we get,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{\dfrac{1-{{e}^{2x}}}{{{e}^{2x}}}}}\]
We get the new expression as,
\[\Rightarrow \dfrac{dy}{dx}=\dfrac{{{e}^{x}}}{\sqrt{1-{{e}^{2x}}}}\]
Therefore, the derivative of the given expression is \[\dfrac{{{e}^{x}}}{\sqrt{1-{{e}^{2x}}}}\].

Note: While writing the differentiation of cosecant function in \[{{\csc }^{-1}}\left( {{e}^{-x}} \right)\], make sure that you take the independent variable as given in the question, which is, the exponential function. Do not take it as \[x\] because then the answer will get wrong. Also, make sure that all computations are done step wise and without any errors.