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Question: What is the derivative of \(e^{-x} \)?...

What is the derivative of e−xe^{-x} ?

Explanation

Solution

We need to solve this function using derivative from first principles, and doing algebra. First substitute this given function into the derivative definition, next by making factors of the numerator by taking out common terms. Next, take e−xe^{-x} out of it and use the exponential function property in limits to find the result. Or else, we can find out by using the chain rule method.

Complete step by step solution:
The formula for derivatives from first principles,
f′(x)=lim⁡h→0f(x+h)−f(x)h...........(1)f'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{f\left( x+h \right)-f\left( x \right)}{h}...........\left( 1 \right)
Let us consider given equation as,
f(x)=e−xf\left( x \right)={{e}^{-x}}
By using equation (1) , we can solve the derivative of given function
Substituting f(x)f\left( x \right)into f′(x)f'\left( x \right),so we get
\Rightarrow $$$$y'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{{{e}^{-\left( x+h \right)}}-{{e}^{-x}}}{h}
Separating the numerator term
\Rightarrow $$$$y'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{{{e}^{-x}}.{{e}^{-h}}-{{e}^{-x}}}{h}
Taking e−xe^{-x} common out in the numerator, then
\Rightarrow $$$$y'\left( x \right)=\displaystyle \lim_{h \to 0}\dfrac{{{e}^{-x}}\left( {{e}^{-h}}-1 \right)}{h}
Here, the e−xe^{-x} is not affected by the limit, so it doesn’t have the ’s values in it and we can factor out of the limit. It gives
\Rightarrow $$$$y'\left( x \right)={{e}^{-x}}\displaystyle \lim_{h \to 0}\dfrac{\left( {{e}^{-h}}-1 \right)}{h}
\Rightarrow $$$$\displaystyle \lim_{h \to 0}\dfrac{\left( {{e}^{-h}}-1 \right)}{h}=-1
Multiplying e−xe^{-x} with -1, then we get
y′(x)=e−x×(−1)y'\left( x \right)={{e}^{-x}}\times \left( -1 \right)
⇒\Rightarrow y′(x)=−e−xy'\left( x \right)=-{{e}^{-x}}
Therefore, the derivative of e−x{{e}^{-x}} is −e−x-{{e}^{-x}}.
Another method is ‘chain rule method’
Given y=e−xy={{e}^{-x}}
Let assume, y=euy={{e}^{u}}
So, u=−x...................(2)u=-x...................\left( 2 \right)
Differentiating the above equation (2) w.r.t ‘y’, it becomes
\Rightarrow $$$$\dfrac{dy}{du}={{e}^{u}}
Again, differentiating equation (2) w.r.t ‘x’,
\Rightarrow $$$$\dfrac{du}{dx}=-1
By using chain rule,
\dfrac{dy}{dx}=\dfrac{dy}{du}.\dfrac{du}{dx}$$$$...........(3)
Substituting the above values into equation (3), then we obtain
\dfrac{dy}{dx}={{e}^{u}}\times \left( -1 \right)$$$$........(4)
\Rightarrow $$$$-{{e}^{u}}
So, we have u=−xu=-x
Then, equation (4) becomes
\Rightarrow $$$$\dfrac{dy}{dx}=-{{e}^{-x}}
Therefore, the derivative of e−x{{e}^{-x}}is −e−x-{{e}^{-x}}.

Note: The common mistakes done by the students is they forget to apply the product or quotient rules. So, remember the derivatives of the product is not the same as applying the product rule. We have two methods for solving, we can solve based on our strategy.