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Differentiation pro !!!
- Thread starter safayah
- Start date
badrobot14
XPRS Administrator
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This is how we did it...
If y = x^x and x > 0 then ln y = ln (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 ' = (ln x + 1)x^x
source
METHOD 2
f(x) = x^x = e^{xln(x)}
when you differentiate e, it stays the same, but you multiply by the inner derrivative:
{df(x)}/{dx} = (ln(x)+1)e^{xln(x)} = (ln(x)+1)x^x
If y = x^x and x > 0 then ln y = ln (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 ' = (ln x + 1)x^x
source
METHOD 2
f(x) = x^x = e^{xln(x)}
when you differentiate e, it stays the same, but you multiply by the inner derrivative:
{df(x)}/{dx} = (ln(x)+1)e^{xln(x)} = (ln(x)+1)x^x
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badrobot14 said:
Thankx for the website !!
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Just because he tried logs doesn't mean he did it right. HallsofIvy way is the correct way to do it. Log both sides then differentiate both.