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Find the derivative of the given term that is \[{x^3}{e^x}\] ?

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Last updated date: 24th Jul 2024
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Answer
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Hint: For differentiation you should know that after differentiating the term or equation given, power of variable is reduced by one, and differentiation of constant is always zero because derivative means measuring the change of a variable with respect to some quantity and as constant is always fixed so no change can be seen.

Complete step-by-step answer:
Given question is
 \[ \Rightarrow \dfrac{d}{{dx}}{x^3}{e^x}\]
To solve this we have to use the product rule of differentiation which states that for any general equation say \[x \times y\] differentiation by product rule can be given as:
 \[ \Rightarrow \dfrac{d}{{dx}}x \times y = y\dfrac{d}{{dx}}x + x\dfrac{d}{{dx}}y\]
Using this formula in our question, on solving we get:
 \[
   \Rightarrow \dfrac{d}{{dx}}{x^3} \times {e^x} = {e^x}\dfrac{d}{{dx}}{x^3} + {x^3}\dfrac{d}{{dx}}{e^x} \\
= {e^x}\left( {3{x^2}} \right) + {x^3} \times {e^x} \times \dfrac{d}{{dx}}\left( x \right) \\
= {e^x}\left( {3{x^2}} \right) + {x^3} \times {e^x} \times (1) \\
   \Rightarrow {e^x}\left( {3{x^2}} \right) + {x^3} \times {e^x} \;
 \]
This is our final answer.
So, the correct answer is “${e^x}\left( {3{x^2}} \right) + {x^3} \times {e^x}$”.

Note: Here in this case only single term was given so it was easy to go through it, but if the equation given is long enough or two variables are given then accordingly you have to solve by assuming one variable as constant and the other one, who is also present in the derivative term should be differentiated.
Differentiation is a very easy concept until and unless the equation given is complicated, in some cases you have to acknowledge the basic properties of differentiation. But in most of the questions you can go on with the basic formulae mentioned above. It's easy to understand once you practice questions over it. Some identities are very much specified like trigonometric identity then in that case you have only the option to learn the direct derivatives.