Question

Let F(x) = e^x, G(x) = e^-x and H(x) = G[F(x)], where x is a real variable. Then ${\displaystyle \frac{\text{dH}}{\text{dx}}}$at x = 0 is ______.

Options

• 1

• - 1

• -${\displaystyle \frac{1}{\text{e}}}$

• - e

Answer 1

Let F(x) = e^x, G(x) = e^-x and H(x) = G[F(x)], where x is a real variable. Then ${\displaystyle \frac{\text{dH}}{\text{dx}}}$at x = 0 is -${\displaystyle \underline{\frac{1}{\text{e}}}}$.

Explanation:

H(x) = G[F(x)]

${\displaystyle ={\text{e}}^{-{\text{e}}^x}}$

∴ H'(x) = ${\displaystyle -{\text{e}}^x\cdot {\text{e}}^{-{\text{e}}^x}}$

∴ H'(0) = ${\displaystyle -{\text{e}}^0\cdot {\text{e}}^{-{\text{e}}^0}}$

${\displaystyle =-{\text{e}}^{-1}}$

${\displaystyle =-\frac{1}{\text{e}}}$