Question

If $y=\ln (e^{mx}+e^{-mx})$, then what is $\frac{dy}{dx}$ at $x=0$ equal to?

• A. $-1$
• B. $0$
• C. $1$
• D. $2$

Answer 1

Answer: (B)

$0$

$y=\ln (e^{mx}+e^{-mx})$

Differentiating with respect to x, we have

$\frac{dy}{dx}=\frac{1}{e^{mx}+e^{-mx}}(m{e}^{mx}-m{e}^{-mx})$

$=0$ .... $($for $x=0)$
Hence, option B.