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If f(x) = x + tan x and f is inverse of g, then g’(x) is equal to
- a)1/1+(g(x)−x)2
- b)1/1−(g(x)−x)2
- c)1/2+(g(x)−x)2
- d)1/2−(g(x)−x)2
Correct answer is option 'C'. Can you explain this answer?
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If f(x) = x + tan x and f is inverse of g, then g’(x) is equal t...
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If f(x) = x + tan x and f is inverse of g, then g’(x) is equal t...
Explanation:
Given:
- f(x) = x + tan x
- f is the inverse of g
Key Concept:
- If f is the inverse of g, then f(g(x)) = x and g(f(x)) = x
Steps to Find g'(x):
1. Start by finding g(x) by replacing f(x) with x in the given equation:
g(x) = x + tan x
2. To find g'(x), differentiate g(x) with respect to x:
g'(x) = 1 + sec^2x
3. Now, we need to express g'(x) in terms of g(x), so substitute g(x) = x + tan x back into g'(x):
g'(x) = 1 + sec^2(x) = 1 + 1 + tan^2x = 2 + tan^2x
4. Rewrite tan^2x as (g(x) - x)^2:
g'(x) = 2 + (g(x) - x)^2 = 2 + (g(x) - x)^2
5. Therefore, g'(x) is equal to 1/2 + (g(x) - x)^2, which corresponds to option C.
Given:
- f(x) = x + tan x
- f is the inverse of g
Key Concept:
- If f is the inverse of g, then f(g(x)) = x and g(f(x)) = x
Steps to Find g'(x):
1. Start by finding g(x) by replacing f(x) with x in the given equation:
g(x) = x + tan x
2. To find g'(x), differentiate g(x) with respect to x:
g'(x) = 1 + sec^2x
3. Now, we need to express g'(x) in terms of g(x), so substitute g(x) = x + tan x back into g'(x):
g'(x) = 1 + sec^2(x) = 1 + 1 + tan^2x = 2 + tan^2x
4. Rewrite tan^2x as (g(x) - x)^2:
g'(x) = 2 + (g(x) - x)^2 = 2 + (g(x) - x)^2
5. Therefore, g'(x) is equal to 1/2 + (g(x) - x)^2, which corresponds to option C.
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If f(x) = x + tan x and f is inverse of g, then g’(x) is equal t...
This question cannot be asked in CA foundation since we do not have derivatives of trigonometric functions
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If f(x) = x + tan x and f is inverse of g, then g’(x) is equal toa)1/1+(g(x)−x)2b)1/1−(g(x)−x)2c)1/2+(g(x)−x)2d)1/2−(g(x)−x)2Correct answer is option 'C'. Can you explain this answer?
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