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$\\frac{d}{dx}(\\ln e^{2x}) = 2$

Question

$\frac{d}{dx}(\ln e^{2x}) = 2$

Explanation:

Step1: Simplify the function

Using the logarithmic property $\ln(ab)=\ln a+\ln b$, we have $\ln(e^{2x})=\ln e + \ln(e^{2x})$. Since $\ln e = 1$ and $\ln(e^{2x})=2x$, the function becomes $1 + 2x$.

Step2: Differentiate the simplified function

Differentiate $y = 1+2x$ with respect to $x$. The derivative of a constant ($1$) is $0$, and the derivative of $2x$ using the power rule $\frac{d}{dx}(ax^n)=anx^{n - 1}$ (here $a = 2$, $n=1$) is $2$. So, $\frac{d}{dx}(1 + 2x)=0+2$.

Answer:

$2$