Find the derivative of y = 2xe^(sqrt[x])

[hank]

Find the derivative of y = 2xe^x^1/2 (y equals 2 times x times e to the square root of x).

My first thought was to use the properties of logs to do this problem. Thus:

lny = ln2 + lnx + ln(e^x^(1/2))
lny = ln2 + lnx + x^(1/2) //because lne is 1.

Here is where I get stuck. My next thought is to "e" everything:

y = 2 + x + e^x^(1/2)

//From here, I take the derivative, but I get the wrong answer.

dy/dx = 1 + e^x^(1/2) * 1/(2x^(1/2))

The answer in the book is: e^x^(1/2) (2 + x^(1/2)).

Help?

 

[stapel]

You have:

. . . . .y = 2xe^sqrt[x]

Taking logs, you get:

. . . . .ln(y) = ln(2xe^sqrt[x])

. . . . .ln(y) = ln(2) + ln(x) + ln(e^sqrt[x])

. . . . .ln(y) = ln(2) + ln(x) + sqrt[x]ln(e)

. . . . .ln(y) = ln(2) + ln(x) + sqrt[x]

Differentiate:

. . . . .(1/y)(dy/dx) = 0 + 1/x + 1/(2sqrt[x])

Multiply through:

. . . . .dy/dx = [1/x + 1/(2sqrt[x])] [y]

. . . . .dy/dx = [1/x + 1/(2sqrt[x])] [2xe^sqrt[x]]

Convert the fractions to a common denominator:

. . . . .dy/dx = [(2 + sqrt[x])/(2x)] [2x e^sqrt[x]]

Complete the simplification.

Eliz.

 

[galactus]

Hello Hankster:

Product rule:

\(\displaystyle \L\\2x\frac{e^{\sqrt{x}}}{2\sqrt{x}}+2e^{\sqrt{x}}\)

\(\displaystyle \L\\\sqrt{x}e^{\sqrt{x}}+2e^{\sqrt{x}}\)

Factor out \(\displaystyle e^{\sqrt{x}}\):

\(\displaystyle \L\\e^{\sqrt{x}}(\sqrt{x}+2)\)

 

[hank]

Ahhh...

So I think I was in the ballpark, I just forgot about substituting back in for y.

Thanks tons!