How would you solve this y', y=x^e e^x?

Question

I'm really confused on the process. I know the answer is e^x x^(e-1) (x+e)
but how do you get e^(x-1)?

Answer 1

y = u^n

y' = n u' u^(n-1)



y = e^u

y' = u'e^u



y' = x^(e-1)e^x + e^x(x^e)

y' = (e^x)( x^(e-1) + x^e )

y' = (e^x)(x^(e-1))(1+1/e)

Answer 2

So to take the derivative, use the product rule:



y = f(x)g(x) --> y' = f'(x)g(x) + f(x)g'(x)

--> here



f(x) = x^e --> f'(x) = e * x^(e - 1)

g(x) = exp(x) --> g'(x) = exp(x)

--> plug into formula for product rule



y' = e * x^(e - 1) * exp(x) + x^e * exp(x)

--> factor out exp(x)



y' = exp(x) * (e * x^(e - 1) + 1 * x^e)

--> x^e = x^(e - 1) * x¹ (since the exponents add to give e - 1 + 1 = e



Now we can factor out x^(e - 1)

-->



y' = exp(x) * x^(e - 1) * (e + 1 * x)

--> or



y' = exp(x) * x^(e - 1) * (x + e)



Edit:



Another way to see that x is leftover when you factor out x^(e - 1) is to just divide by that factor:



e * x^(e - 1) + x^e = x^(e - 1) * { e * x^(e - 1) / x^(e - 1) + x^e / x^(e - 1) }

--> just divide everything by what you factored out!



well in the first part x^(e - 1) directly cancels leaving ONLY e...in the second, let's write it out:



x^e / x^(e - 1) = x^e / (x^e * x⁻¹)

--> x^e's cancel, leaving:



1 / x⁻¹ = 1 / (1/x)

--> multiply by reciprocal:



1 * x/1 = x --> x^e / x^(e - 1) = x



So this is why x and e are leftover...you COULD have factored out just e^x, but then you would get:



y' = exp(x) * x^e * (e/x + 1)



(because now x⁻¹ would be left from doing x^(e - 1) / x^e = 1/x)

Answer 3

For this, we'll need to product rule, which states that the derivative of a function f(x) = u*v is equal to u*dv + v*du. The derivative of x^e is equal to e*x^(e-1), by the derivative rule for polynomials. The derivative of e^x is equal to e^x, by the derivative rule for exponents. Thus, the derivative is (x^e)*(e^x) + (e^x)*(e*x^(e-1)). You can factor and combine like terms to get (e^x)(x^e + e*x^(e-1)), from which you can also factor out x^(e-1) to get (e^x)*(x^(e-1))*(x+e)