Question

question 5 of 26
\\(\frac{d}{dx}\tan e^{-x}=\\)
-\\(e^{-x}\sec^{2}e^{-x}\\)
\\(e^{-x}\sec^{2}e^{-x}\\)
\\(\sec^{2}e^{-x}\\)
-\\(x\sec^{2}e^{-x}\\)

Explanation:

Step1: Apply chain - rule

The derivative of $\tan(u)$ with respect to $x$ is $\sec^{2}(u)\frac{du}{dx}$. Here $u = e^{-x}$.

Step2: Find derivative of $u = e^{-x}$

The derivative of $e^{-x}$ with respect to $x$ is $-e^{-x}$ using the rule that the derivative of $e^{ax}$ with respect to $x$ is $ae^{ax}$ (where $a=- 1$).

Step3: Combine results

$\frac{d}{dx}\tan(e^{-x})=\sec^{2}(e^{-x})\times(-e^{-x})=-e^{-x}\sec^{2}(e^{-x})$

Answer:

$-e^{-x}\sec^{2}(e^{-x})$