If f(x)={x²+1}, g(x)={x+1}{x²+1}, h(x)=2 x-3, then f^{}≤ft(h^{}≤ft(g^{}(x)))=

Correct Answer is : 25{2}{{5}}52

Correct Answer is : 25{2}{{5}}52

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MHTCET 13 Apr 2026 Shift 2 Paper

Question : 122 of 150

Marks: +1, -0

f(x)=\sqrt{x^{2}+1}, g(x)=\frac{x+1}{x^{2}+1}, h(x)=2 x-3

f^{\prime}\left(h^{\prime}\left(g^{\prime}(x)\right)\right)=

Solution:

Concept:

Use differentiation, then evaluate the composed derivatives step by step.

Explanation:

Given

h(x)=2x-3

Differentiate:

h'(x)=2

Since

h'(x)

is constantly

2

, for any input we get

h'\bigl(g'(x)\bigr)=2

Now find

f'(x)

f(x)=\sqrt{x^2+1}=(x^2+1)^{1/2}

Using the chain rule:

f'(x)=\frac{1}{2}(x^2+1)^{-1/2}\cdot 2x=\frac{x}{\sqrt{x^2+1}}

We need

f'(h'(g'(x)))=f'(2)

Substitute

x=2

f'(2)=\frac{2}{\sqrt{2^2+1}}=\frac{2}{\sqrt{5}}

Answer:

\frac{2}{\sqrt{5}}

i.e. Option C.

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