Let f(x) = e^x , g(x) = ^-1 x and h(x) = f(g(x)) , then h'(x)/h(x) is: - Mathematics MCQ Mathematics Mock Test - 9 2026 | ExamTest.live

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Let f (x) = e^{x}, g (x) = sin^{- 1} x and h (x) = f (g (x)), then \frac{h ^{'} ( x )}{h ( x )} is:

Solution & Step-by-step Explanation

Given f (x) = e^{x} and g (x) = sin^{- 1} x . h (x) = f (g (x)) = e^{s i n^{- 1} x} .Differentiating h (x) with respect to x using the chain rule: h^{'} (x) = \frac{d}{d x} (e^{s i n^{- 1} x}) = e^{s i n^{- 1} x} \cdot \frac{d}{d x} (sin^{- 1} x) h^{'} (x) = e^{s i n^{- 1} x} \cdot \frac{1}{1 - x ^{2}} Now, find the ratio: \frac{h ^{'} ( x )}{h ( x )} = \frac{e ^{s i n^{- 1} x} \cdot \frac{1}{1 - x ^{2}}}{e ^{s i n^{- 1} x}} = \frac{1}{1 - x ^{2}}

Try it yourself before checking the explanation above.

Let f (x) = e^{x}, g (x) = sin^{- 1} x and h (x) = f (g (x)), then \frac{h ^{'} ( x )}{h ( x )} is:

e^{s i n^{- 1} x}

\frac{1}{1 - x ^{2}}

sin^{- 1} x

\frac{1}{1 - x ^{2}}

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