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Mathematics Question on Differentiability

Let F(x)=ex,G(x)=e−xF(x) = e^x, G(x) = e^{-x} and H(x)=G(F(x))H(x) = G(F(x)), where xx is a real variable. Then dHdx\frac{dH}{dx} at x=0x = 0 is

A

11

B

−1-1

C

−1e-\frac{1}{e}

D

−e-e

Answer

−1e-\frac{1}{e}

Explanation

Solution

We have, F(x)=ex,G(x)=e−xF(x)=e^{x}, G(x)=e^{-x}
∴H(x)=G(F(x))\therefore H(x) =G(F(x))
=G(ex)=G\left(e^{x}\right)
=e−ex=e^{-e^{x}}
∴dHdx=e−ex⋅(−ex)\therefore \frac{d H}{d x} =e^{-e^{x}} \cdot\left(-e^{x}\right)
=−exe−ex=-e^{x} e^{-e^{x}}
∴dHdx∣x=0=−e0⋅e−e∘\therefore \left.\frac{d H}{d x}\right|_{x=0} =-e^{0} \cdot e^{-e^{\circ}}
=−1⋅e−1=-1 \cdot e^{-1}
=−e−1=-e^{-1}
=−1e=\frac{-1}{e}