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(d/dx){log(secx+tanx)}=
- a)cos x
- b)sec x
- c)tan x
- d)cot x
Correct answer is option 'B'. Can you explain this answer?
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Verified Answer
(d/dx){log(secx+tanx)}=a)cos xb)sec xc)tan xd)cot xCorrect answer is o...
d d x { log S e c x + tan x }
= 1 S e c x + tan x S e c x . tan x + sec 2 x
= S e c x S e c x + tan x S e c x + tan x
= S e c x
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(d/dx){log(secx+tanx)}=a)cos xb)sec xc)tan xd)cot xCorrect answer is o...
Explanation:
Given Function: \(f(x) = \log(\sec x + \tan x)\)
Derivative of the Function: \(\frac{d}{dx}[\log(\sec x + \tan x)]\)
Using Chain Rule:
- Let \(u = \sec x + \tan x\)
- \(f(x) = \log(u)\)
- \(\frac{df}{dx} = \frac{1}{u} \cdot \frac{du}{dx}\)
Finding \(\frac{du}{dx}\):
- \(\frac{du}{dx} = \frac{d}{dx}[\sec x + \tan x]\)
- \(\frac{du}{dx} = \sec x \tan x + \sec^2 x\)
- \(\frac{du}{dx} = \sec x(\sec x + \tan x)\)
Substitute \(\frac{du}{dx}\) back into the derivative:
- \(\frac{df}{dx} = \frac{1}{\sec x + \tan x} \cdot \sec x(\sec x + \tan x)\)
- \(\frac{df}{dx} = \sec x\)
Therefore, the derivative of the given function is \(\sec x\), which corresponds to option B.
Given Function: \(f(x) = \log(\sec x + \tan x)\)
Derivative of the Function: \(\frac{d}{dx}[\log(\sec x + \tan x)]\)
Using Chain Rule:
- Let \(u = \sec x + \tan x\)
- \(f(x) = \log(u)\)
- \(\frac{df}{dx} = \frac{1}{u} \cdot \frac{du}{dx}\)
Finding \(\frac{du}{dx}\):
- \(\frac{du}{dx} = \frac{d}{dx}[\sec x + \tan x]\)
- \(\frac{du}{dx} = \sec x \tan x + \sec^2 x\)
- \(\frac{du}{dx} = \sec x(\sec x + \tan x)\)
Substitute \(\frac{du}{dx}\) back into the derivative:
- \(\frac{df}{dx} = \frac{1}{\sec x + \tan x} \cdot \sec x(\sec x + \tan x)\)
- \(\frac{df}{dx} = \sec x\)
Therefore, the derivative of the given function is \(\sec x\), which corresponds to option B.
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(d/dx){log(secx+tanx)}=a)cos xb)sec xc)tan xd)cot xCorrect answer is o...
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(d/dx){log(secx+tanx)}=a)cos xb)sec xc)tan xd)cot xCorrect answer is option 'B'. Can you explain this answer?
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