Question

How do you differentiate \[y = x{e^x}\] ?

Answer 1

Hint: Here in this to differentiate given function by using the product rule of differentiation, which states that \[\dfrac{d}{{dx}}\left[ {f(x).g(x)} \right] = f(x).\dfrac{d}{{dx}}g(x) + g(x).\dfrac{d}{{dx}}f(x)\] where f(x) and g(x) is a function of x, further on simplification differentiate the using power and exponential rule .

Complete step-by-step answer:
Differentiation is the algebraic method of finding the derivative for a function at any point. The derivative is a concept that is at the root of calculus. ... Either way, both the slope and the instantaneous rate of change are equivalent, and the function to find both of these at any point is called the derivative. It can be represented as \[\dfrac{{dy}}{{dx}}\] , where y is the function of x.
Consider the given function
 \[ \Rightarrow \,\,\,\,y = x{e^x}\]
In the given question, you can see that \[x{e^x}\] is a product of \[x\] and \[{e^x}\] , both which are elementary functions, then
Differentiate with respect to x using the product rule i.e., \[\dfrac{d}{{dx}}\left[ {f(x).g(x)} \right] = f(x).\dfrac{d}{{dx}}g(x) + g(x).\dfrac{d}{{dx}}f(x)\]
Where \[f(x) = x\] and \[g(x) = {e^x}\] , then
 \[ \Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = \dfrac{d}{{dx}}\left( {x{e^x}} \right) = x.\dfrac{d}{{dx}}{e^x} + {e^x}.\dfrac{d}{{dx}}x\]
Differentiate using the exponential rule which states that \[\dfrac{d}{{dx}}\left[ {{a^x}} \right] = {a^x}\ln \left( a \right)\] where a=e i.e., \[\dfrac{d}{{dx}}\left( {{e^x}} \right) = {e^x}\]
 \[ \Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = x.{e^x} + {e^x}.\dfrac{d}{{dx}}x\]
Differentiate using the power rule i.e., \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\]
 \[ \Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = x.{e^x} + {e^x}.1\]
 \[ \Rightarrow \,\,\,\,\dfrac{{dy}}{{dx}} = x{e^x} + {e^x}\]
Take \[{e^x}\] as common in RHS
 \[\therefore \,\,\,\,\dfrac{{dy}}{{dx}} = {e^x}\left( {x + 1} \right)\]
Hence, the differentiated term of a function \[y = x{e^x}\] is \[\,\,\,\dfrac{{dy}}{{dx}} = {e^x}\left( {x + 1} \right)\] .
So, the correct answer is “ $ {e^x}\left( {x + 1} \right) $ ”.

Note: The differentiation is defined as the derivative of a function with respect to the independent variable. Here the dependent variable is y and the independent variable is x. If the function is a product of more than one function, we use product rule to find the derivative. The product rule is defined as \[\dfrac{d}{{dx}}(uv) = u\dfrac{{dv}}{{dx}} + v\dfrac{{du}}{{dx}}\] , where u and v are both the function of x. By using the differentiation formulas we can obtain the result.