Find the derivative of y with respect to the given independent variable

Question

Find the derivative of y with respect to the given independent variable:

$y = 3^{-x} \stackrel{D}{\longrightarrow} y' = 3^{-x} \cdot (-1) \cdot \ln 3 $

This is my teacher's solution. I don't understand what he did. Can someone explain to me?

Answer 1

Your teacher implicitly wrote the function $y$ as \begin{eqnarray} y &=& 3^{-x} \\ &=& e^{-x \ln 3} \end{eqnarray} From which the derivative is straightforward \begin{eqnarray} \frac{dy}{dx} &=& - \ln 3 . e^{-x \ln 3} \\ &=& (-1).3^{-x}.\ln 3 \end{eqnarray}

Answer 2

HINT to solve:

\begin{equation} y=3^{-x} => e^{\ln(3^{-x})} => e^{(-x)\ln(3)} \end{equation}

Answer 3

Suppose that you have $$y=a^{f(x)}$$ Take logarithms of both sides $$\log(y)=f(x) \log(a)$$ Exponentiate $$y=e^{f(x) \log(a)}$$ Now, compute the derivative $$y'=f'(x)\log(a)e^{f(x) \log(a)}$$ Replace again $e^{f(x) \log(a)}$ by $a^{f(x)}$ and get $$y'=f'(x)\log(a) a^{f(x)}$$

I am sure that you can take from here.

Answer 4

use this standard result$$\frac{d a^x}{dx}= a^xln(a)$$ and apply chain rule