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How do you find the derivative of ${e^{ - 3x}}$?

Answer
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Hint:
Here we need to know that the derivative of $\dfrac{d}{{dx}}{e^{ax}} = a{e^{ax}}$ which means that the coefficient of the $x$ in the power needs to be noticed and terms with the $x$ in the power of the exponential are also needed to be multiplied with the term which we have got.

Complete step by step solution:
Here we are given to find the derivative of ${e^{ax}}$
We need to know that when we have the derivative term as ${e^x}$ then its differentiation is also the same ${e^x}$.
But here there is not simply the term ${e^x}$instead we have the term where we also have the coefficient in the power with the variable $x$ so we need to know that when we have the coefficient of degree then that coefficient will also be multiplied with the differentiation of the exponential.
$\dfrac{d}{{dx}}{e^{ax}} = a{e^{ax}}$
So here also we have multiplied the differentiation of the coefficient of the $x$along with the exponential term after the differentiation.
$\dfrac{d}{{dx}}{e^{ax}} = a{e^{ax}}$$ - - - (1)$
Now comparing ${e^{ax}}$ with${e^{ - 3x}}$, we can write that:
$a = - 3$
Now we know that differentiation of $\dfrac{d}{{dx}}{e^{ax}} = a{e^{ax}}$
Hence we can compare and say:
$\dfrac{d}{{dx}}{e^{ - 3x}}$
So we will get $\dfrac{d}{{dx}}{e^{ - 3x}} = {e^{ - 3x}}.\dfrac{d}{{dx}}\left( { - 3x} \right)$$ - - - - - (2)$
We also know that differentiation of $\dfrac{d}{{dx}}\left( {ax} \right) = a$
Hence in the similar way we can say that:
$\dfrac{d}{{dx}}\left( { - 3x} \right) = - 3$
So now we need to substitute this value in the equation (2) and we will get:
$\dfrac{d}{{dx}}{e^{ - 3x}} = {e^{ - 3x}}.\dfrac{d}{{dx}}\left( { - 3x} \right) = {e^{ - 3x}}.\left( { - 3} \right)$

$\dfrac{d}{{dx}}{e^{ - 3x}} = - 3{e^{ - 3x}}$

Note:
Here the student must know that whenever we need to differentiate we also need to multiply with exponential terms the differentiation of the power of $e$.
Similarly when we need to integrate then we need to divide the term instead of multiplication.