Question

Let F(x) = e^x, G(x) = e^–x& H(x) = G(F(x)) where x is a real

number then $\frac{dH}{dx}$at x = 0 is –

• A. 1
• B. –1
• C. – 1/e
• D. – e

Answer 1

Answer: (C)

– 1/e

Answer 2

H(x) = G(F(x)) = G(e^x) = $e^{–e^{x}}$

$\frac{dH(x)}{dx}$ = ${e^{–e}}^{x}$. (– e^x)

$\left( \frac{dH(x)}{dx} \right)_{x = 0}$ = ${e^{–e}}^{0}$.(– e⁰) = e^–1.(–1) = – $\frac{1}{e}$