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Question: How do you find the derivative of \(y = 2{e^x}\) ?...

How do you find the derivative of y=2exy = 2{e^x} ?

Explanation

Solution

Derivative of this function can be found in two ways:
By using the ‘First principle of derivative’, which is also a definition of derivative of a function.
Already knowing the derivative of ex{e^x}.
Here, let us use the first method.

Formula used:
f′(x)=lim⁡h→0f(x+h)−f(x)hf\prime (x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}
Where h>0h > 0 and is a constant.
Also,
lim⁡x→0ex−1x=1\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - 1}}{x} = 1

Complete step-by-step answer:
We are given y=f(x)=2exy = f(x) = 2{e^x}
Therefore, f(x+h)=2e(x+h)f(x + h) = 2{e^{(x + h)}}
The equation of derivative is:
f′(x)=lim⁡h→0f(x+h)−f(x)hf\prime (x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{f(x + h) - f(x)}}{h}
Now, we have to substitute the values in the above equation to obtain
f′(x)=lim⁡h→02e(x+h)−2exhf\prime (x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{2{e^{(x + h)}} - 2{e^x}}}{h}
⇒f′(x)=lim⁡h→02ex×eh−2exh\Rightarrow f\prime (x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{2{e^x} \times {e^h} - 2{e^x}}}{h}
⇒f′(x)=lim⁡h→02ex(eh−1)h\Rightarrow f\prime (x) = \mathop {\lim }\limits_{h \to 0} \dfrac{{2{e^x}({e^h} - 1)}}{h}
⇒f′(x)=2ex×lim⁡h→0eh−1h\Rightarrow f\prime (x) = 2{e^x} \times \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^h} - 1}}{h} , since the limit depends on the variable hh and not on the variable xx .
Now we know that,
lim⁡x→0ex−1x=1\mathop {\lim }\limits_{x \to 0} \dfrac{{{e^x} - 1}}{x} = 1 , which is a standard limit result.
⇒f′(x)=2ex×1\Rightarrow f\prime (x) = 2{e^x} \times 1
⇒f′(x)=2ex\Rightarrow f\prime (x) = 2{e^x}
Thus, the derivative of our function y=2exy = 2{e^x} is 2ex2{e^x}.

Additional information:
Using the first principle of derivatives to find derivatives is tedious and time consuming. It is, therefore, better to know some standard derivatives which makes these kinds of problems easier.
For example, here we could use the result that the derivative of ex{e^x}is ex{e^x}. Then, we must know that ‘derivative of a function multiplied by a constant is equal to that constant times derivative of the function’. The second part is a result from the product rule of differentiation. Using this two information, solving this equation becomes a lot more easier and time conserving.

Note: Standard limit values should be known for using the first principle of derivatives while standard derivative results should be known to solve otherwise (using properties of derivatives).