Approximating derivatives using $sin(x)$ and $cos(x)$
$begingroup$
In a question I plotted an approximated derivative of $sin(x)$ at the point $frac{pi}{4}$. After that I plotted the absolute error in this derivative approximation. I used $cos(x)$ as the exact solution.
My question is will the exact solution of an approximated derivative of $sin(x)$ always be $cos(x)$? And then if it was the other way around and I was approximating $cos(x)$ would I use $-sin(x)$ as its exact solution?
calculus algebra-precalculus
asked Dec 4 '18 at 15:31
mt12345mt12345
958
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add a comment |
$begingroup$
In a question I plotted an approximated derivative of $sin(x)$ at the point $frac{pi}{4}$. After that I plotted the absolute error in this derivative approximation. I used $cos(x)$ as the exact solution.
My question is will the exact solution of an approximated derivative of $sin(x)$ always be $cos(x)$? And then if it was the other way around and I was approximating $cos(x)$ would I use $-sin(x)$ as its exact solution?
calculus algebra-precalculus
asked Dec 4 '18 at 15:31
mt12345mt12345
958
$endgroup$
1
$begingroup$
Yes, $sin' x = cos x$ and $cos' x = -sin x$. (These are exact.)
$endgroup$
– KM101
Dec 4 '18 at 15:33
add a comment |
$begingroup$
In a question I plotted an approximated derivative of $sin(x)$ at the point $frac{pi}{4}$. After that I plotted the absolute error in this derivative approximation. I used $cos(x)$ as the exact solution.
My question is will the exact solution of an approximated derivative of $sin(x)$ always be $cos(x)$? And then if it was the other way around and I was approximating $cos(x)$ would I use $-sin(x)$ as its exact solution?
calculus algebra-precalculus
asked Dec 4 '18 at 15:31
mt12345mt12345
958
$endgroup$
In a question I plotted an approximated derivative of $sin(x)$ at the point $frac{pi}{4}$. After that I plotted the absolute error in this derivative approximation. I used $cos(x)$ as the exact solution.
My question is will the exact solution of an approximated derivative of $sin(x)$ always be $cos(x)$? And then if it was the other way around and I was approximating $cos(x)$ would I use $-sin(x)$ as its exact solution?
calculus algebra-precalculus
calculus algebra-precalculus
asked Dec 4 '18 at 15:31
mt12345mt12345
958
asked Dec 4 '18 at 15:31
mt12345mt12345
958
asked Dec 4 '18 at 15:31
mt12345mt12345
958
asked Dec 4 '18 at 15:31
mt12345mt12345
958
asked Dec 4 '18 at 15:31
mt12345mt12345
958
958
1
$begingroup$
Yes, $sin' x = cos x$ and $cos' x = -sin x$. (These are exact.)
$endgroup$
– KM101
Dec 4 '18 at 15:33
add a comment |
1
$begingroup$
Yes, $sin' x = cos x$ and $cos' x = -sin x$. (These are exact.)
$endgroup$
– KM101
Dec 4 '18 at 15:33
1
1
$begingroup$
Yes, $sin' x = cos x$ and $cos' x = -sin x$. (These are exact.)
$endgroup$
– KM101
Dec 4 '18 at 15:33
$begingroup$
Yes, $sin' x = cos x$ and $cos' x = -sin x$. (These are exact.)
$endgroup$
– KM101
Dec 4 '18 at 15:33
add a comment |
1 Answer
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$begingroup$
$frac{d}{dx}sin x=cos x$
$frac{d}{dx}cos x=-sin x$
These are indeed the exact "solutions".
answered Dec 4 '18 at 15:38
Thomas ShelbyThomas Shelby
2,408221
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$begingroup$
$frac{d}{dx}sin x=cos x$
$frac{d}{dx}cos x=-sin x$
These are indeed the exact "solutions".
answered Dec 4 '18 at 15:38
Thomas ShelbyThomas Shelby
2,408221
$endgroup$
add a comment |
$begingroup$
$frac{d}{dx}sin x=cos x$
$frac{d}{dx}cos x=-sin x$
These are indeed the exact "solutions".
answered Dec 4 '18 at 15:38
Thomas ShelbyThomas Shelby
2,408221
$endgroup$
add a comment |
$begingroup$
$frac{d}{dx}sin x=cos x$
$frac{d}{dx}cos x=-sin x$
These are indeed the exact "solutions".
answered Dec 4 '18 at 15:38
Thomas ShelbyThomas Shelby
2,408221
$endgroup$
$frac{d}{dx}sin x=cos x$
$frac{d}{dx}cos x=-sin x$
These are indeed the exact "solutions".
answered Dec 4 '18 at 15:38
Thomas ShelbyThomas Shelby
2,408221
answered Dec 4 '18 at 15:38
Thomas ShelbyThomas Shelby
2,408221
answered Dec 4 '18 at 15:38
Thomas ShelbyThomas Shelby
2,408221
answered Dec 4 '18 at 15:38
Thomas ShelbyThomas Shelby
2,408221
2,408221
add a comment |
add a comment |
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$begingroup$
Yes, $sin' x = cos x$ and $cos' x = -sin x$. (These are exact.)
$endgroup$
– KM101
Dec 4 '18 at 15:33