Question
If f(x) = x + tan x and f is inverse of g, then g’(x) is equal to
Answer: Option C
:
C
Lety=f(x)⇒x=f−1(y)
thenf(x)=x+tanx
⇒x=f−1(y)+tan(f−1(y))
⇒x=g(y)+tan(g(y))orx=g(x)+tan(g(x))⋯(i)
Differentiating both sides, then we get
1=g′(x)+sec2g(x).g′(x)
∴g′(x)=11+sec2(g(x))=11+1+tan2(g(x))
=12+(x−g(x))2[fromEq.(i)]
12+(g(x)−x)2
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:
C
Lety=f(x)⇒x=f−1(y)
thenf(x)=x+tanx
⇒x=f−1(y)+tan(f−1(y))
⇒x=g(y)+tan(g(y))orx=g(x)+tan(g(x))⋯(i)
Differentiating both sides, then we get
1=g′(x)+sec2g(x).g′(x)
∴g′(x)=11+sec2(g(x))=11+1+tan2(g(x))
=12+(x−g(x))2[fromEq.(i)]
12+(g(x)−x)2
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