How do you find the derivative of \[2{e^{ - x}}\]?

Hint: Derivatives are defined as the varying rate of a function with respect to an independent variable. Since they ask us to find only differentiation of ‘y’ with respect to ‘x’. Since we have constant 2 multiplied to \[{e^{ - x}}\]. We take the constant outside the derivative and then we differentiate it.

Complete step-by-step solution:
Given, \[2{e^{ - x}}\].
Let take \[y = 2{e^{ - x}}\]
Now differentiating with respect to ‘x’. We have
\[\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}(2{e^{ - x}})\]
\[\Rightarrow \dfrac{d}{{dx}}(2{e^{ - x}})\]
Now taking constant 2 outside the derivative,
\[\Rightarrow 2\dfrac{d}{{dx}}({e^{ - x}})\]
We know that the differentiation of \[{e^x}\] is \[{e^x}\]. But we have power as ‘-x’.
\[\Rightarrow 2 \times {e^{ - x}}\dfrac{{d( - x)}}{{dx}}\]
Here we have negative 1 multiplied to ‘x’. taking negative sign outside we have
\[\Rightarrow - 2 \times {e^{ - x}}\dfrac{{dx}}{{dx}}\]
\[\Rightarrow - 2 \times {e^{ - x}}\]

Hence differentiation of \[2{e^{ - x}}\] is \[ - 2 \times {e^{ - x}}\]

Additional information:
i) Linear combination rule: The linearity law is very important to emphasize its nature with alternate notation. Symbolically it is specified as \[h'(x) = af'(x) + bg'(x)\]
ii) Quotient rule: The derivative of one function divided by other is found by quotient rule such as\[{\left[ {\dfrac{{f(x)}}{{g(x)}}} \right]^{'}} = \dfrac{{g(x)f{'}(x) - f(x)g{'}(x)}}{{{{\left[ {g(x)} \right]}^2}}}\].
iii) Product rule: When a derivative of a product of two function is to be found, then we use product rule that is \[\dfrac{{dy}}{{dx}} = u \times \dfrac{{dv}}{{dx}} + v \times \dfrac{{du}}{{dx}}\].
iv) Chain rule: To find the derivative of composition function or function of a function, we use chain rule. That is \[fog'({x_0}) = [(f'og)({x_0})]g'({x_0})\].

Note: We know the differentiation of \[{x^n}\] is \[\dfrac{{d({x^n})}}{{dx}} = n.{x^{n - 1}}\]. The obtained result is the first derivative. If we differentiate again we get a second derivative. If we differentiate the second derivative again we get a third derivative and so on. Careful in applying the product rule. We also know that differentiation of constant terms is zero.