
Find \[f'\left( x \right)\].
\[f\left( x \right)=\sec x-\sqrt{2}\tan x\]
Answer
574.2k+ views
Hint: To solve the above problem first we have to find the basic derivatives of \[\sec x\] and \[\tan x\]. After substituting the derivatives in the equation, rewrite the equation with the derivatives of the function. Solve the equation to find the final answer.
Complete step-by-step answer:
Applying derivative on both sides of the equation with respect to x we get,
\[f'\left( x \right)=\dfrac{d}{dx}\left( \sec x \right)-\dfrac{d}{dx}\left( \sqrt{2}\tan x \right)\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know the derivative of \[\sec x\] is \[\sec x\cdot \tan x\] and the derivative of \[\tan x\] is \[{{\sec }^{2}}x\].
On substituting the derivatives of \[\sec x\] and \[\tan x\] in the above equation we get,
\[f'\left( x \right)=\sec x\cdot \tan x-\sqrt{2}{{\sec }^{2}}x\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Taking \[\sec x\] as common in the right hand side (RHS) we get,
\[f'\left( x \right)=\sec x\left( \tan x-\sqrt{2}\sec x \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
Hence the value of \[f'\left( x \right)\] is \[\sec x\left( \tan x-\sqrt{2}\sec x \right)\].
Note: The possible error that you may encounter can be the wrong substitution values of the derivatives of \[\sec x\] and \[\tan x\]. Solving the equation should be done carefully. It is to note here that integers are exempted from the calculation of derivatives.
Complete step-by-step answer:
Applying derivative on both sides of the equation with respect to x we get,
\[f'\left( x \right)=\dfrac{d}{dx}\left( \sec x \right)-\dfrac{d}{dx}\left( \sqrt{2}\tan x \right)\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (1)
We know the derivative of \[\sec x\] is \[\sec x\cdot \tan x\] and the derivative of \[\tan x\] is \[{{\sec }^{2}}x\].
On substituting the derivatives of \[\sec x\] and \[\tan x\] in the above equation we get,
\[f'\left( x \right)=\sec x\cdot \tan x-\sqrt{2}{{\sec }^{2}}x\] . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Taking \[\sec x\] as common in the right hand side (RHS) we get,
\[f'\left( x \right)=\sec x\left( \tan x-\sqrt{2}\sec x \right)\]. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
Hence the value of \[f'\left( x \right)\] is \[\sec x\left( \tan x-\sqrt{2}\sec x \right)\].
Note: The possible error that you may encounter can be the wrong substitution values of the derivatives of \[\sec x\] and \[\tan x\]. Solving the equation should be done carefully. It is to note here that integers are exempted from the calculation of derivatives.
Recently Updated Pages
Master Class 12 English: Engaging Questions & Answers for Success

Master Class 12 Social Science: Engaging Questions & Answers for Success

Master Class 12 Chemistry: Engaging Questions & Answers for Success

Which is the Longest Railway Platform in the world?

India Manned Space Mission Launch Target Month and Year 2025 Update

Which of the following pairs is correct?

Trending doubts
What are the major means of transport Explain each class 12 social science CBSE

Which are the Top 10 Largest Countries of the World?

Draw a labelled sketch of the human eye class 12 physics CBSE

How much time does it take to bleed after eating p class 12 biology CBSE

Explain sex determination in humans with line diag class 12 biology CBSE

Plot a graph between potential difference V and current class 12 physics CBSE

