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In Leibniz notation, how do you write the second derivative of y with respect to the square of x?
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I know how to write in Leibniz notation for more plain-vanilla expressions like the the second derivative of y with respect to x. But I am not sure how to write, in Leibniz notation, the second derivative of y with respect to expressions more complex than a single variable. In, for example, the second derivative of y with respect to the square of x, would I just use parentheses and write $dy^2/d(x^2)^2$?
calculus derivatives notation
edited Jan 7 at 18:50
gt6989b
33.9k22455
asked Jan 7 at 18:48
resplaineresplaine
61
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add a comment |
$begingroup$
I know how to write in Leibniz notation for more plain-vanilla expressions like the the second derivative of y with respect to x. But I am not sure how to write, in Leibniz notation, the second derivative of y with respect to expressions more complex than a single variable. In, for example, the second derivative of y with respect to the square of x, would I just use parentheses and write $dy^2/d(x^2)^2$?
calculus derivatives notation
edited Jan 7 at 18:50
gt6989b
33.9k22455
asked Jan 7 at 18:48
resplaineresplaine
61
$endgroup$
1
$begingroup$
It’s $d^2y$, not $dy^2$.
$endgroup$
– KM101
Jan 7 at 18:50
add a comment |
$begingroup$
I know how to write in Leibniz notation for more plain-vanilla expressions like the the second derivative of y with respect to x. But I am not sure how to write, in Leibniz notation, the second derivative of y with respect to expressions more complex than a single variable. In, for example, the second derivative of y with respect to the square of x, would I just use parentheses and write $dy^2/d(x^2)^2$?
calculus derivatives notation
edited Jan 7 at 18:50
gt6989b
33.9k22455
asked Jan 7 at 18:48
resplaineresplaine
61
$endgroup$
I know how to write in Leibniz notation for more plain-vanilla expressions like the the second derivative of y with respect to x. But I am not sure how to write, in Leibniz notation, the second derivative of y with respect to expressions more complex than a single variable. In, for example, the second derivative of y with respect to the square of x, would I just use parentheses and write $dy^2/d(x^2)^2$?
calculus derivatives notation
calculus derivatives notation
edited Jan 7 at 18:50
gt6989b
33.9k22455
asked Jan 7 at 18:48
resplaineresplaine
61
edited Jan 7 at 18:50
gt6989b
33.9k22455
asked Jan 7 at 18:48
resplaineresplaine
61
edited Jan 7 at 18:50
gt6989b
33.9k22455
edited Jan 7 at 18:50
gt6989b
33.9k22455
edited Jan 7 at 18:50
gt6989b
33.9k22455
33.9k22455
asked Jan 7 at 18:48
resplaineresplaine
61
asked Jan 7 at 18:48
resplaineresplaine
61
asked Jan 7 at 18:48
resplaineresplaine
61
61
1
$begingroup$
It’s $d^2y$, not $dy^2$.
$endgroup$
– KM101
Jan 7 at 18:50
add a comment |
1
$begingroup$
It’s $d^2y$, not $dy^2$.
$endgroup$
– KM101
Jan 7 at 18:50
1
1
$begingroup$
It’s $d^2y$, not $dy^2$.
$endgroup$
– KM101
Jan 7 at 18:50
$begingroup$
It’s $d^2y$, not $dy^2$.
$endgroup$
– KM101
Jan 7 at 18:50
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Indeed, if $z=x^2$ and you need
$$
frac{d^2y}{dz^2} = frac{d^2 y}{dleft(x^2right)^2}
$$
answered Jan 7 at 18:50
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
Thanks so much! That's just the answer I needed.
$endgroup$
– resplaine
Jan 7 at 18:58
add a comment |
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1 Answer
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active
oldest
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Indeed, if $z=x^2$ and you need
$$
frac{d^2y}{dz^2} = frac{d^2 y}{dleft(x^2right)^2}
$$
answered Jan 7 at 18:50
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
Thanks so much! That's just the answer I needed.
$endgroup$
– resplaine
Jan 7 at 18:58
add a comment |
$begingroup$
Indeed, if $z=x^2$ and you need
$$
frac{d^2y}{dz^2} = frac{d^2 y}{dleft(x^2right)^2}
$$
answered Jan 7 at 18:50
gt6989bgt6989b
33.9k22455
$endgroup$
$begingroup$
Thanks so much! That's just the answer I needed.
$endgroup$
– resplaine
Jan 7 at 18:58
add a comment |
$begingroup$
Indeed, if $z=x^2$ and you need
$$
frac{d^2y}{dz^2} = frac{d^2 y}{dleft(x^2right)^2}
$$
answered Jan 7 at 18:50
gt6989bgt6989b
33.9k22455
$endgroup$
Indeed, if $z=x^2$ and you need
$$
frac{d^2y}{dz^2} = frac{d^2 y}{dleft(x^2right)^2}
$$
answered Jan 7 at 18:50
gt6989bgt6989b
33.9k22455
answered Jan 7 at 18:50
gt6989bgt6989b
33.9k22455
answered Jan 7 at 18:50
gt6989bgt6989b
33.9k22455
answered Jan 7 at 18:50
gt6989bgt6989b
33.9k22455
33.9k22455
$begingroup$
Thanks so much! That's just the answer I needed.
$endgroup$
– resplaine
Jan 7 at 18:58
add a comment |
$begingroup$
Thanks so much! That's just the answer I needed.
$endgroup$
– resplaine
Jan 7 at 18:58
$begingroup$
Thanks so much! That's just the answer I needed.
$endgroup$
– resplaine
Jan 7 at 18:58
$begingroup$
Thanks so much! That's just the answer I needed.
$endgroup$
– resplaine
Jan 7 at 18:58
add a comment |
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1
$begingroup$
It’s $d^2y$, not $dy^2$.
$endgroup$
– KM101
Jan 7 at 18:50