Let $F(x) = e^x, G(x) = e^{-x}$ and $H(x) = G(F(x))$, where $x$ is a real variable. Then $\frac{dH}{dx}$ at $x = 0$ is
- $1$
- $-1$
- $-\frac{1}{e}$
- $-e$
The Correct Option is C
Solution and Explanation
$\therefore H(x) =G(F(x)) $
$=G\left(e^{x}\right) $
$=e^{-e^{x}} $
$\therefore \frac{d H}{d x} =e^{-e^{x}} \cdot\left(-e^{x}\right) $
$=-e^{x} e^{-e^{x}} $
$\therefore \left.\frac{d H}{d x}\right|_{x=0} =-e^{0} \cdot e^{-e^{\circ}}$
$=-1 \cdot e^{-1}$
$=-e^{-1}$
$=\frac{-1}{e}$
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Concepts Used:
Continuity
A function is said to be continuous at a point x = a, if
limx→a
f(x) Exists, and
limx→a
f(x) = f(a)
It implies that if the left hand limit (L.H.L), right hand limit (R.H.L) and the value of the function at x=a exists and these parameters are equal to each other, then the function f is said to be continuous at x=a.
If the function is undefined or does not exist, then we say that the function is discontinuous.
Conditions for continuity of a function: For any function to be continuous, it must meet the following conditions:
- The function f(x) specified at x = a, is continuous only if f(a) belongs to real number.
- The limit of the function as x approaches a, exists.
- The limit of the function as x approaches a, must be equal to the function value at x = a.