
Differentiate the term $\tan {e^{ - y}}$.
Answer
581.4k+ 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.
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
What is meant by exothermic and endothermic reactions class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

What are Quantum numbers Explain the quantum number class 11 chemistry CBSE

