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Question: Find \[f'\left( x \right)\]. \[f\left( x \right)=\sec x-\sqrt{2}\tan x\]...

Find f(x)f'\left( x \right).
f(x)=secx2tanxf\left( x \right)=\sec x-\sqrt{2}\tan x

Explanation

Solution

Hint : To solve the above problem first we have to find the basic derivatives of secx\sec x and tanx\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(x)=ddx(secx)ddx(2tanx)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 secx\sec x is secxtanx\sec x\cdot \tan x and the derivative of tanx\tan x is sec2x{{\sec }^{2}}x.
On substituting the derivatives of secx\sec x and tanx\tan x in the above equation we get,
f(x)=secxtanx2sec2xf'\left( x \right)=\sec x\cdot \tan x-\sqrt{2}{{\sec }^{2}}x . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (2)
Taking secx\sec x as common in the right hand side (RHS) we get,
f(x)=secx(tanx2secx)f'\left( x \right)=\sec x\left( \tan x-\sqrt{2}\sec x \right). . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (3)
Hence the value of f(x)f'\left( x \right) is secx(tanx2secx)\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 secx\sec x and tanx\tan x. Solving the equation should be done carefully. It is to note here that integers are exempted from the calculation of derivatives.