$x=frac{1}{4}(t+e^{-t})$, $y=e^{-frac{t}{2}}$, prove that $yy''(x)=(y'(x))^2cdot sqrt{1+(y'(x))^2}$: what am...
$begingroup$
$x=frac{1}{4}(t+e^{-t})$, $y=e^{-frac{t}{2}}$, prove that
$yy''(x)=(y'(x))^2cdot sqrt{1+(y'(x))^2}$
$$yy''(x)=e^{-frac{t}{2}}cdot frac{(e^{-t/2})'' (frac{1}{4}(t+e^{-t})'-(frac{1}{4}(t+e^{-t}))''(e^{-t/2})'}{((e^{-t/2})')^3}$$
Derivating and simplifying, I get
$$-frac{1}{2}(sqrt{e^t}+frac{1}{sqrt{e^t}}).$$
Which is a negative value, and can't be equal to what the exercise requires.
I am very sorry I didn't type out my 'interim calculations': MathJax is time-comsuming, it's late, and there's still homework left.
What I need to know, was there an arithmetical mistake, or do I misunderstand something fundamentally?
Did I understand the exercise incorrectly? Did I get the 
wrong somewhere?
Thank you.
calculus derivatives
asked Nov 26 '18 at 13:07
fragileradiusfragileradius
297114
$endgroup$
add a comment |
$begingroup$
$x=frac{1}{4}(t+e^{-t})$, $y=e^{-frac{t}{2}}$, prove that
$yy''(x)=(y'(x))^2cdot sqrt{1+(y'(x))^2}$
$$yy''(x)=e^{-frac{t}{2}}cdot frac{(e^{-t/2})'' (frac{1}{4}(t+e^{-t})'-(frac{1}{4}(t+e^{-t}))''(e^{-t/2})'}{((e^{-t/2})')^3}$$
Derivating and simplifying, I get
$$-frac{1}{2}(sqrt{e^t}+frac{1}{sqrt{e^t}}).$$
Which is a negative value, and can't be equal to what the exercise requires.
I am very sorry I didn't type out my 'interim calculations': MathJax is time-comsuming, it's late, and there's still homework left.
What I need to know, was there an arithmetical mistake, or do I misunderstand something fundamentally?
Did I understand the exercise incorrectly? Did I get the
wrong somewhere?
Thank you.
calculus derivatives
asked Nov 26 '18 at 13:07
fragileradiusfragileradius
297114
$endgroup$
$begingroup$
Wait, I think I might have put $y$ into the denominator instead of $x$...
$endgroup$
– fragileradius
Nov 26 '18 at 13:15
add a comment |
$begingroup$
$x=frac{1}{4}(t+e^{-t})$, $y=e^{-frac{t}{2}}$, prove that
$yy''(x)=(y'(x))^2cdot sqrt{1+(y'(x))^2}$
$$yy''(x)=e^{-frac{t}{2}}cdot frac{(e^{-t/2})'' (frac{1}{4}(t+e^{-t})'-(frac{1}{4}(t+e^{-t}))''(e^{-t/2})'}{((e^{-t/2})')^3}$$
Derivating and simplifying, I get
$$-frac{1}{2}(sqrt{e^t}+frac{1}{sqrt{e^t}}).$$
Which is a negative value, and can't be equal to what the exercise requires.
I am very sorry I didn't type out my 'interim calculations': MathJax is time-comsuming, it's late, and there's still homework left.
What I need to know, was there an arithmetical mistake, or do I misunderstand something fundamentally?
Did I understand the exercise incorrectly? Did I get the
wrong somewhere?
Thank you.
calculus derivatives
asked Nov 26 '18 at 13:07
fragileradiusfragileradius
297114
$endgroup$
$x=frac{1}{4}(t+e^{-t})$, $y=e^{-frac{t}{2}}$, prove that
$yy''(x)=(y'(x))^2cdot sqrt{1+(y'(x))^2}$
$$yy''(x)=e^{-frac{t}{2}}cdot frac{(e^{-t/2})'' (frac{1}{4}(t+e^{-t})'-(frac{1}{4}(t+e^{-t}))''(e^{-t/2})'}{((e^{-t/2})')^3}$$
Derivating and simplifying, I get
$$-frac{1}{2}(sqrt{e^t}+frac{1}{sqrt{e^t}}).$$
Which is a negative value, and can't be equal to what the exercise requires.
I am very sorry I didn't type out my 'interim calculations': MathJax is time-comsuming, it's late, and there's still homework left.
What I need to know, was there an arithmetical mistake, or do I misunderstand something fundamentally?
Did I understand the exercise incorrectly? Did I get the
wrong somewhere?
Thank you.
calculus derivatives
calculus derivatives
asked Nov 26 '18 at 13:07
fragileradiusfragileradius
297114
asked Nov 26 '18 at 13:07
fragileradiusfragileradius
297114
asked Nov 26 '18 at 13:07
fragileradiusfragileradius
297114
asked Nov 26 '18 at 13:07
fragileradiusfragileradius
297114
asked Nov 26 '18 at 13:07
fragileradiusfragileradius
297114
297114
$begingroup$
Wait, I think I might have put $y$ into the denominator instead of $x$...
$endgroup$
– fragileradius
Nov 26 '18 at 13:15
add a comment |
$begingroup$
Wait, I think I might have put $y$ into the denominator instead of $x$...
$endgroup$
– fragileradius
Nov 26 '18 at 13:15
$begingroup$
Wait, I think I might have put $y$ into the denominator instead of $x$...
$endgroup$
– fragileradius
Nov 26 '18 at 13:15
$begingroup$
Wait, I think I might have put $y$ into the denominator instead of $x$...
$endgroup$
– fragileradius
Nov 26 '18 at 13:15
add a comment |
1 Answer
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$begingroup$
My mistake, I have put $y$ into the denominator instead of $x$.
answered Dec 2 '18 at 4:33
fragileradiusfragileradius
297114
$endgroup$
add a comment |
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$begingroup$
My mistake, I have put $y$ into the denominator instead of $x$.
answered Dec 2 '18 at 4:33
fragileradiusfragileradius
297114
$endgroup$
add a comment |
$begingroup$
My mistake, I have put $y$ into the denominator instead of $x$.
answered Dec 2 '18 at 4:33
fragileradiusfragileradius
297114
$endgroup$
add a comment |
$begingroup$
My mistake, I have put $y$ into the denominator instead of $x$.
answered Dec 2 '18 at 4:33
fragileradiusfragileradius
297114
$endgroup$
My mistake, I have put $y$ into the denominator instead of $x$.
answered Dec 2 '18 at 4:33
fragileradiusfragileradius
297114
answered Dec 2 '18 at 4:33
fragileradiusfragileradius
297114
answered Dec 2 '18 at 4:33
fragileradiusfragileradius
297114
answered Dec 2 '18 at 4:33
fragileradiusfragileradius
297114
297114
add a comment |
add a comment |
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$begingroup$
Wait, I think I might have put $y$ into the denominator instead of $x$...
$endgroup$
– fragileradius
Nov 26 '18 at 13:15