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Finding the second derivative of $x^x$
The 2019 Stack Overflow Developer Survey Results Are InOn the meaning of the second derivativeHelp finding the second derivative of this function.Derivative of function composed with itselfSecond Derivative Of A Parametric FunctionSecond derivative of the cost function of logistic functionProduct Rule and Differentiation with logarithmsDerivative of matrix determinant wrt to matrix elementHow to take second derivative implicitlyFind $alpha in mathbbR$ s.t. the second derivative ($x=0$) of a function exists.Derivative of the von Neumann entropy
$begingroup$
Find the second derivative $d^2y/dx^2$ when $y=x^x:(x>0)$.
$$y=x^x,::(xgt0)$$
beginalign
log y&=xlog x \
rmDifferentiating&:rmwith:respect:to:x
endalign
beginalign
frac1yfracdydx&=1cdot(log x+1)+xcdotfrac1x \[0.8ex]
fracdydx&=x^x(log x+1)
endalign
I found the first derivative, and now I want to know how to find the second derivative of this function.
derivatives
edited Mar 23 at 4:51
Robert Howard
2,3033935
asked Jun 10 '16 at 5:12
Aakash KumarAakash Kumar
2,5341520
$endgroup$
add a comment |
$begingroup$
Find the second derivative $d^2y/dx^2$ when $y=x^x:(x>0)$.
$$y=x^x,::(xgt0)$$
beginalign
log y&=xlog x \
rmDifferentiating&:rmwith:respect:to:x
endalign
beginalign
frac1yfracdydx&=1cdot(log x+1)+xcdotfrac1x \[0.8ex]
fracdydx&=x^x(log x+1)
endalign
I found the first derivative, and now I want to know how to find the second derivative of this function.
derivatives
edited Mar 23 at 4:51
Robert Howard
2,3033935
asked Jun 10 '16 at 5:12
Aakash KumarAakash Kumar
2,5341520
$endgroup$
$begingroup$
Well, you know what the derivative of $x^x$ is, and you know what the derivative of ($logx + 1$) is, so just use the product rule.
$endgroup$
– Aurey
Jun 10 '16 at 5:15
$begingroup$
Hint: Product rule.
$endgroup$
– user223391
Jun 10 '16 at 5:15
add a comment |
$begingroup$
Find the second derivative $d^2y/dx^2$ when $y=x^x:(x>0)$.
$$y=x^x,::(xgt0)$$
beginalign
log y&=xlog x \
rmDifferentiating&:rmwith:respect:to:x
endalign
beginalign
frac1yfracdydx&=1cdot(log x+1)+xcdotfrac1x \[0.8ex]
fracdydx&=x^x(log x+1)
endalign
I found the first derivative, and now I want to know how to find the second derivative of this function.
derivatives
edited Mar 23 at 4:51
Robert Howard
2,3033935
asked Jun 10 '16 at 5:12
Aakash KumarAakash Kumar
2,5341520
$endgroup$
Find the second derivative $d^2y/dx^2$ when $y=x^x:(x>0)$.
$$y=x^x,::(xgt0)$$
beginalign
log y&=xlog x \
rmDifferentiating&:rmwith:respect:to:x
endalign
beginalign
frac1yfracdydx&=1cdot(log x+1)+xcdotfrac1x \[0.8ex]
fracdydx&=x^x(log x+1)
endalign
I found the first derivative, and now I want to know how to find the second derivative of this function.
derivatives
derivatives
edited Mar 23 at 4:51
Robert Howard
2,3033935
asked Jun 10 '16 at 5:12
Aakash KumarAakash Kumar
2,5341520
edited Mar 23 at 4:51
Robert Howard
2,3033935
asked Jun 10 '16 at 5:12
Aakash KumarAakash Kumar
2,5341520
edited Mar 23 at 4:51
Robert Howard
2,3033935
edited Mar 23 at 4:51
Robert Howard
2,3033935
edited Mar 23 at 4:51
Robert Howard
2,3033935
2,3033935
asked Jun 10 '16 at 5:12
Aakash KumarAakash Kumar
2,5341520
asked Jun 10 '16 at 5:12
Aakash KumarAakash Kumar
2,5341520
asked Jun 10 '16 at 5:12
Aakash KumarAakash Kumar
2,5341520
2,5341520
$begingroup$
Well, you know what the derivative of $x^x$ is, and you know what the derivative of ($logx + 1$) is, so just use the product rule.
$endgroup$
– Aurey
Jun 10 '16 at 5:15
$begingroup$
Hint: Product rule.
$endgroup$
– user223391
Jun 10 '16 at 5:15
add a comment |
$begingroup$
Well, you know what the derivative of $x^x$ is, and you know what the derivative of ($logx + 1$) is, so just use the product rule.
$endgroup$
– Aurey
Jun 10 '16 at 5:15
$begingroup$
Hint: Product rule.
$endgroup$
– user223391
Jun 10 '16 at 5:15
$begingroup$
Well, you know what the derivative of $x^x$ is, and you know what the derivative of ($logx + 1$) is, so just use the product rule.
$endgroup$
– Aurey
Jun 10 '16 at 5:15
$begingroup$
Well, you know what the derivative of $x^x$ is, and you know what the derivative of ($logx + 1$) is, so just use the product rule.
$endgroup$
– Aurey
Jun 10 '16 at 5:15
$begingroup$
Hint: Product rule.
$endgroup$
– user223391
Jun 10 '16 at 5:15
$begingroup$
Hint: Product rule.
$endgroup$
– user223391
Jun 10 '16 at 5:15
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Just do the same thing: $$log left( fracdydx right) = x log x + log (log x + 1),$$ so $$frac1fracdydx fracd^2 ydx^2 = (log x + 1) + frac1log x + 1 cdot frac1x,$$ hence $$fracd^2 ydx^2 = fracdydx left( log x + 1 + frac1x(log x + 1) right),$$ and substitute the expression you obtained for the first derivative.
Alternatively, $$fracdydx = x^x (log x + 1)$$ implies $$fracd^2 ydx^2 = fracddxleft[x^x right] (log x + 1) + x^x cdot frac1x$$ by the product rule, and as before, substitute the derivative of $x^x$ that you found earlier.
answered Jun 10 '16 at 5:20
heropupheropup
65.4k865104
$endgroup$
add a comment |
$begingroup$
$y = x^x$
As shown, $fracdydx = x^x(log x + 1)$
Differentiating, again w.r.t. $x$, we get,
$frac d^2ydx^2 = fracddxx^x (log x + 1)$
Using given information and differentiating by parts, we get,
$$frac d^2ydx^2 = (log x + 1)fracddxx^x + x^x fracddx(log x + 1)$$
$$frac d^2ydx^2 = (log x + 1).x^x(log x + 1) + x^x(frac 1x + 0)$$
$$frac d^2ydx^2 = x^x(log x + 1)^2 + x^x-1$$
Required answer.
answered Jun 10 '16 at 5:28
Shikhar S. MaheshwariShikhar S. Maheshwari
308112
$endgroup$
add a comment |
$begingroup$
This is not an answer but it is too long for a comment.
You have been already given good answers.
What could be interesting is to think how $y$, $y'$, $y''$ would have be coded in the most inexpensive way. $$colorredy=x^ximplies log(y)=x log(x)$$ so $$fracy'y=1+log (x)implies colorredy'=yleft(1+log (x) right)implieslog(y')=log(y)+log (1+log (x))$$ $$fracy''y'=fracy'y+frac1x (1+log (x))implies colorredy''=y'left(fracy'y+frac1x (1+log (x))right)$$ As you can see, very few basic operations (we could compute $log(x)$ only once and reuse it)
answered Jun 10 '16 at 6:32
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Just do the same thing: $$log left( fracdydx right) = x log x + log (log x + 1),$$ so $$frac1fracdydx fracd^2 ydx^2 = (log x + 1) + frac1log x + 1 cdot frac1x,$$ hence $$fracd^2 ydx^2 = fracdydx left( log x + 1 + frac1x(log x + 1) right),$$ and substitute the expression you obtained for the first derivative.
Alternatively, $$fracdydx = x^x (log x + 1)$$ implies $$fracd^2 ydx^2 = fracddxleft[x^x right] (log x + 1) + x^x cdot frac1x$$ by the product rule, and as before, substitute the derivative of $x^x$ that you found earlier.
answered Jun 10 '16 at 5:20
heropupheropup
65.4k865104
$endgroup$
add a comment |
$begingroup$
Just do the same thing: $$log left( fracdydx right) = x log x + log (log x + 1),$$ so $$frac1fracdydx fracd^2 ydx^2 = (log x + 1) + frac1log x + 1 cdot frac1x,$$ hence $$fracd^2 ydx^2 = fracdydx left( log x + 1 + frac1x(log x + 1) right),$$ and substitute the expression you obtained for the first derivative.
Alternatively, $$fracdydx = x^x (log x + 1)$$ implies $$fracd^2 ydx^2 = fracddxleft[x^x right] (log x + 1) + x^x cdot frac1x$$ by the product rule, and as before, substitute the derivative of $x^x$ that you found earlier.
answered Jun 10 '16 at 5:20
heropupheropup
65.4k865104
$endgroup$
add a comment |
$begingroup$
Just do the same thing: $$log left( fracdydx right) = x log x + log (log x + 1),$$ so $$frac1fracdydx fracd^2 ydx^2 = (log x + 1) + frac1log x + 1 cdot frac1x,$$ hence $$fracd^2 ydx^2 = fracdydx left( log x + 1 + frac1x(log x + 1) right),$$ and substitute the expression you obtained for the first derivative.
Alternatively, $$fracdydx = x^x (log x + 1)$$ implies $$fracd^2 ydx^2 = fracddxleft[x^x right] (log x + 1) + x^x cdot frac1x$$ by the product rule, and as before, substitute the derivative of $x^x$ that you found earlier.
answered Jun 10 '16 at 5:20
heropupheropup
65.4k865104
$endgroup$
Just do the same thing: $$log left( fracdydx right) = x log x + log (log x + 1),$$ so $$frac1fracdydx fracd^2 ydx^2 = (log x + 1) + frac1log x + 1 cdot frac1x,$$ hence $$fracd^2 ydx^2 = fracdydx left( log x + 1 + frac1x(log x + 1) right),$$ and substitute the expression you obtained for the first derivative.
Alternatively, $$fracdydx = x^x (log x + 1)$$ implies $$fracd^2 ydx^2 = fracddxleft[x^x right] (log x + 1) + x^x cdot frac1x$$ by the product rule, and as before, substitute the derivative of $x^x$ that you found earlier.
answered Jun 10 '16 at 5:20
heropupheropup
65.4k865104
answered Jun 10 '16 at 5:20
heropupheropup
65.4k865104
answered Jun 10 '16 at 5:20
heropupheropup
65.4k865104
answered Jun 10 '16 at 5:20
heropupheropup
65.4k865104
65.4k865104
add a comment |
add a comment |
$begingroup$
$y = x^x$
As shown, $fracdydx = x^x(log x + 1)$
Differentiating, again w.r.t. $x$, we get,
$frac d^2ydx^2 = fracddxx^x (log x + 1)$
Using given information and differentiating by parts, we get,
$$frac d^2ydx^2 = (log x + 1)fracddxx^x + x^x fracddx(log x + 1)$$
$$frac d^2ydx^2 = (log x + 1).x^x(log x + 1) + x^x(frac 1x + 0)$$
$$frac d^2ydx^2 = x^x(log x + 1)^2 + x^x-1$$
Required answer.
answered Jun 10 '16 at 5:28
Shikhar S. MaheshwariShikhar S. Maheshwari
308112
$endgroup$
add a comment |
$begingroup$
$y = x^x$
As shown, $fracdydx = x^x(log x + 1)$
Differentiating, again w.r.t. $x$, we get,
$frac d^2ydx^2 = fracddxx^x (log x + 1)$
Using given information and differentiating by parts, we get,
$$frac d^2ydx^2 = (log x + 1)fracddxx^x + x^x fracddx(log x + 1)$$
$$frac d^2ydx^2 = (log x + 1).x^x(log x + 1) + x^x(frac 1x + 0)$$
$$frac d^2ydx^2 = x^x(log x + 1)^2 + x^x-1$$
Required answer.
answered Jun 10 '16 at 5:28
Shikhar S. MaheshwariShikhar S. Maheshwari
308112
$endgroup$
add a comment |
$begingroup$
$y = x^x$
As shown, $fracdydx = x^x(log x + 1)$
Differentiating, again w.r.t. $x$, we get,
$frac d^2ydx^2 = fracddxx^x (log x + 1)$
Using given information and differentiating by parts, we get,
$$frac d^2ydx^2 = (log x + 1)fracddxx^x + x^x fracddx(log x + 1)$$
$$frac d^2ydx^2 = (log x + 1).x^x(log x + 1) + x^x(frac 1x + 0)$$
$$frac d^2ydx^2 = x^x(log x + 1)^2 + x^x-1$$
Required answer.
answered Jun 10 '16 at 5:28
Shikhar S. MaheshwariShikhar S. Maheshwari
308112
$endgroup$
$y = x^x$
As shown, $fracdydx = x^x(log x + 1)$
Differentiating, again w.r.t. $x$, we get,
$frac d^2ydx^2 = fracddxx^x (log x + 1)$
Using given information and differentiating by parts, we get,
$$frac d^2ydx^2 = (log x + 1)fracddxx^x + x^x fracddx(log x + 1)$$
$$frac d^2ydx^2 = (log x + 1).x^x(log x + 1) + x^x(frac 1x + 0)$$
$$frac d^2ydx^2 = x^x(log x + 1)^2 + x^x-1$$
Required answer.
answered Jun 10 '16 at 5:28
Shikhar S. MaheshwariShikhar S. Maheshwari
308112
answered Jun 10 '16 at 5:28
Shikhar S. MaheshwariShikhar S. Maheshwari
308112
answered Jun 10 '16 at 5:28
Shikhar S. MaheshwariShikhar S. Maheshwari
308112
answered Jun 10 '16 at 5:28
Shikhar S. MaheshwariShikhar S. Maheshwari
308112
308112
add a comment |
add a comment |
$begingroup$
This is not an answer but it is too long for a comment.
You have been already given good answers.
What could be interesting is to think how $y$, $y'$, $y''$ would have be coded in the most inexpensive way. $$colorredy=x^ximplies log(y)=x log(x)$$ so $$fracy'y=1+log (x)implies colorredy'=yleft(1+log (x) right)implieslog(y')=log(y)+log (1+log (x))$$ $$fracy''y'=fracy'y+frac1x (1+log (x))implies colorredy''=y'left(fracy'y+frac1x (1+log (x))right)$$ As you can see, very few basic operations (we could compute $log(x)$ only once and reuse it)
answered Jun 10 '16 at 6:32
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
This is not an answer but it is too long for a comment.
You have been already given good answers.
What could be interesting is to think how $y$, $y'$, $y''$ would have be coded in the most inexpensive way. $$colorredy=x^ximplies log(y)=x log(x)$$ so $$fracy'y=1+log (x)implies colorredy'=yleft(1+log (x) right)implieslog(y')=log(y)+log (1+log (x))$$ $$fracy''y'=fracy'y+frac1x (1+log (x))implies colorredy''=y'left(fracy'y+frac1x (1+log (x))right)$$ As you can see, very few basic operations (we could compute $log(x)$ only once and reuse it)
answered Jun 10 '16 at 6:32
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
This is not an answer but it is too long for a comment.
You have been already given good answers.
What could be interesting is to think how $y$, $y'$, $y''$ would have be coded in the most inexpensive way. $$colorredy=x^ximplies log(y)=x log(x)$$ so $$fracy'y=1+log (x)implies colorredy'=yleft(1+log (x) right)implieslog(y')=log(y)+log (1+log (x))$$ $$fracy''y'=fracy'y+frac1x (1+log (x))implies colorredy''=y'left(fracy'y+frac1x (1+log (x))right)$$ As you can see, very few basic operations (we could compute $log(x)$ only once and reuse it)
answered Jun 10 '16 at 6:32
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
This is not an answer but it is too long for a comment.
You have been already given good answers.
What could be interesting is to think how $y$, $y'$, $y''$ would have be coded in the most inexpensive way. $$colorredy=x^ximplies log(y)=x log(x)$$ so $$fracy'y=1+log (x)implies colorredy'=yleft(1+log (x) right)implieslog(y')=log(y)+log (1+log (x))$$ $$fracy''y'=fracy'y+frac1x (1+log (x))implies colorredy''=y'left(fracy'y+frac1x (1+log (x))right)$$ As you can see, very few basic operations (we could compute $log(x)$ only once and reuse it)
answered Jun 10 '16 at 6:32
Claude LeiboviciClaude Leibovici
125k1158135
answered Jun 10 '16 at 6:32
Claude LeiboviciClaude Leibovici
125k1158135
answered Jun 10 '16 at 6:32
Claude LeiboviciClaude Leibovici
125k1158135
answered Jun 10 '16 at 6:32
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
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$begingroup$
Well, you know what the derivative of $x^x$ is, and you know what the derivative of ($logx + 1$) is, so just use the product rule.
$endgroup$
– Aurey
Jun 10 '16 at 5:15
$begingroup$
Hint: Product rule.
$endgroup$
– user223391
Jun 10 '16 at 5:15