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Calculus Level 2

Find the derivative of the function $y = x^x.$

y'=x^x (\ln x + 1)

y'=x\ln x

y'=x (\ln x + 1)

y'=x^{x+1} (\ln x + 1)

3 solutions

Alex Kasantsidis
May 27, 2015

$y=x^{x}$

$lny=xlnx$

$\frac{dy}{dx} lny=\frac{dy}{dx} xlnx$

using implicit differentiation you get

$\frac{1}{y} \frac{dy}{dx}= lnx+1$

$\frac{dy}{dx}=y(lnx+1)$

recall $y=x^{x}$

therefore

$\frac{dy}{dx} = x^{x}(lnx+1)$

Moderator note:

Yes, this is the standard logarithmic differentiation approach. Bonus question: Do we still need to apply logarithmic differentiation if we want to determine $\frac {d^2 y}{dx^2}, \frac{d^3y}{dx^3} ,\ldots$ and so on?

Answer to challenge note: Not necessary. We already know the first derivative of x^x and can re-use that

Anand Santhanakrishnan - 6 years ago

I found the same here.

Marcelo Dias Jr. - 6 years ago

Shouldn't the third line be

$\frac{d}{dx}lny=\frac{d}{dx}xlnx$

? then

$\frac{d}{dy}lny\frac{dy}{dx}=lnx+1$

Ka Kei Yeung - 6 years ago

i skipped that step and just took the derivate of lny with respect to y and showed it which was 1/y

Alex Kasantsidis - 6 years ago

I'm not saying you skipped any step, I'm saying you wrote dy instead of d

Ka Kei Yeung - 6 years ago

@Ka Kei Yeung – Yes, because that is implicit differentiation. when you take the derivate of lny you use the chain rule and take the derivate of y with respect to x which is 1 times dy/dx

Alex Kasantsidis - 6 years ago

@Alex Kasantsidis – " Yes, because that is implicit differentiation. when you take the derivate of lny you use the chain rule and take the derivate of y with respect to x which is 1 times dy/dx" Yes by chain rule:

$\frac{d}{dx}lny=\frac{d}{dx}xlnx$

$\frac{d}{dy}lny\frac{dy}{dx}=lnx+1$

$\frac{1}{y}\frac{dy}{dx}=lnx + 1$

Ka Kei Yeung - 6 years ago

@Ka Kei Yeung – So i just skipped your 3rd step. I am sorry.

Alex Kasantsidis - 6 years ago

@Alex Kasantsidis – No the skipping is fine, but not the dy instead of d in YOUR 3rd line.

Ka Kei Yeung - 6 years ago

Max Beech
May 26, 2015

Refer to image

Moderator note:

Correct. Can you type it out here? It's not entirely legible.

$y={ x }^{ x }\\ y={ e }^{ ln({ x }^{ x }) }\\ y={ e }^{ xln(x) }\\ y'=(ln(x)+1)({ e }^{ xln(x) })\\ y'=(ln(x)+1){ x }^{ x }$

Max Beech - 6 years ago

I like your method. However, did you forget parenthesis around the "ln(x) + 1"?

Louis W - 6 years ago

Ahah yes Louis thanks, I got lazy there! Greetings from the UK!

Max Beech - 6 years ago

Mahasvin Gogi
Aug 20, 2015

If y = x x

Use properties of logarithmic functions to expand the right side of the above equation as follows.

ln y = x ln x

We now differentiate both sides with respect to x, using chain rule on the left side and the product rule on the right.

y '(1 / y) = ln x + x(1 / x) = ln x + 1 , where y ' = dy/dx

Multiply both sides by y

y ' = (ln x + 1)y

Substitute y by x ^x to obtain

y ' = x^x(ln x + 1)

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