
Differentiate the term $\tan {e^{ - y}}$.
Answer
595.2k+ views
Hint: In order to differentiate the term $\tan {e^{ - y}}$ we need to know that the differentiation of tan x with respect to x is ${\sec ^2}x$ and that of ${e^x}$ with respect to x is ${e^x}$ and $ - x$ with respect to x is -1. Here we have to first differentiate tan then e then –y with respect to y to get the right answer.
Complete step by step answer:
The given term which is being differentiated is $\tan {e^{ - y}}$.
Let P = $\tan {e^{ - y}}$.
On differentiating P with respect to y we get,
$
\dfrac{{dP}}{{dy}} = \dfrac{{d\left( {\tan {e^{ - y}}} \right)}}{{dy}} \\
\dfrac{{dP}}{{dy}} = {\sec ^2}{e^{ - y}}\left( {\dfrac{{d{e^{ - y}}}}{{dy}}} \right) \\
\dfrac{{dP}}{{dy}} = {e^{ - y}}{\sec ^2}{e^{ - y}}\left( {\dfrac{{d\left( { - y} \right)}}{{dy}}} \right) \\
\dfrac{{dP}}{{dy}} = - {e^{ - y}}{\sec ^2}{e^{ - y}} \\
$
As we know that the differentiation of tan x with respect to x is ${\sec ^2}x$ and that of ${e^x}$ with respect to x is ${e^x}$ and $ - x$ with respect to x is -1. According to this we have proceeded above and differentiated the term with respect to y.
So, the differentiation of the term $\tan {e^{ - y}}$ with respect to y is $ - {e^{ - y}}{\sec ^2}{e^{ - y}}$
So, the answer to this problem is $ - {e^{ - y}}{\sec ^2}{e^{ - y}}$.
Note: For such problems of differentiation we need to know chain rule. In chain rule we have to differentiate every term separately and then we have to write it together. You need to know that differentiation is a process of finding a function that outputs the rate of change of one variable with respect to another variable. Knowing this will solve your problem.
Complete step by step answer:
The given term which is being differentiated is $\tan {e^{ - y}}$.
Let P = $\tan {e^{ - y}}$.
On differentiating P with respect to y we get,
$
\dfrac{{dP}}{{dy}} = \dfrac{{d\left( {\tan {e^{ - y}}} \right)}}{{dy}} \\
\dfrac{{dP}}{{dy}} = {\sec ^2}{e^{ - y}}\left( {\dfrac{{d{e^{ - y}}}}{{dy}}} \right) \\
\dfrac{{dP}}{{dy}} = {e^{ - y}}{\sec ^2}{e^{ - y}}\left( {\dfrac{{d\left( { - y} \right)}}{{dy}}} \right) \\
\dfrac{{dP}}{{dy}} = - {e^{ - y}}{\sec ^2}{e^{ - y}} \\
$
As we know that the differentiation of tan x with respect to x is ${\sec ^2}x$ and that of ${e^x}$ with respect to x is ${e^x}$ and $ - x$ with respect to x is -1. According to this we have proceeded above and differentiated the term with respect to y.
So, the differentiation of the term $\tan {e^{ - y}}$ with respect to y is $ - {e^{ - y}}{\sec ^2}{e^{ - y}}$
So, the answer to this problem is $ - {e^{ - y}}{\sec ^2}{e^{ - y}}$.
Note: For such problems of differentiation we need to know chain rule. In chain rule we have to differentiate every term separately and then we have to write it together. You need to know that differentiation is a process of finding a function that outputs the rate of change of one variable with respect to another variable. Knowing this will solve your problem.
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