How would I go about differentiating t with respect to x? The 2019 Stack Overflow Developer Survey Results Are In Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Need help differentiating this equationsDifferentiating with fractionsDifferentiating an implicit equation.Differentiating a function with respect to two unknown.Differentiating a Triangle Wave function?Derivative with respect to aDifferentiating both sides of an equation with different variablesDifferentiating with respect to the exponent of distributiondifferentiating equations with expectationsDifferentiating with respect to a product of two variables
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How would I go about differentiating t with respect to x?
The 2019 Stack Overflow Developer Survey Results Are In
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Need help differentiating this equationsDifferentiating with fractionsDifferentiating an implicit equation.Differentiating a function with respect to two unknown.Differentiating a Triangle Wave function?Derivative with respect to aDifferentiating both sides of an equation with different variablesDifferentiating with respect to the exponent of distributiondifferentiating equations with expectationsDifferentiating with respect to a product of two variables
$begingroup$
$$t=fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3)$$
We haven't covered how to differentiate equations like this yet. How would I go about finding dt/dx?
derivatives
asked Mar 24 at 16:00
methyl0x_methyl0x_
103
$endgroup$
add a comment |
$begingroup$
$$t=fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3)$$
We haven't covered how to differentiate equations like this yet. How would I go about finding dt/dx?
derivatives
asked Mar 24 at 16:00
methyl0x_methyl0x_
103
$endgroup$
add a comment |
$begingroup$
$$t=fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3)$$
We haven't covered how to differentiate equations like this yet. How would I go about finding dt/dx?
derivatives
asked Mar 24 at 16:00
methyl0x_methyl0x_
103
$endgroup$
$$t=fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3)$$
We haven't covered how to differentiate equations like this yet. How would I go about finding dt/dx?
derivatives
derivatives
asked Mar 24 at 16:00
methyl0x_methyl0x_
103
asked Mar 24 at 16:00
methyl0x_methyl0x_
103
edited Mar 24 at 16:33
methyl0x_
asked Mar 24 at 16:00
methyl0x_methyl0x_
103
asked Mar 24 at 16:00
methyl0x_methyl0x_
103
asked Mar 24 at 16:00
methyl0x_methyl0x_
103
103
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Here's a sneaky way around some of the work others have suggested. In particular, you don't need to know how to differentiate the square-root function.
Do a little rewriting via algebra:
beginalign
t&=fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3)\
t-fracxv &=fracsqrt(fraca2)^2+(a-x)^2(fracv3)\
fracv3left(t-fracxvright) &=sqrt(fraca2)^2+(a-x)^2\
left(fracv3right)^2left(t-fracxvright)^2 &=(fraca2)^2+(a-x)^2
endalign
Now remembering that $t$ is really $t(x)$, a function that varies with $x$, we have a equality between two functions -- one on the left, one on the right. So their derivatives must be equal. The one on the left needs the power rule and the chain rule; the one on the right needs only the power rule. Differentiating, we get
beginalign
2left(fracv3right)^2left(t-fracxvright)left(fracdtdx + frac-1vright) &=2(a-x)(-1)\
left(fracv3right)^2left(t-fracxvright)left(fracdtdx + frac-1vright) &=(x-a)\
left(t-fracxvright)left(fracdtdx + frac-1vright) &=(x-a)left(frac3vright)^2\
fracdtdx + frac-1v &=frac(x-a)left(frac3vright)^2t-fracxv\
fracdtdx &=frac(x-a)left(frac3vright)^2t-fracxv + frac1v\
endalign
At this point, you have a formula for $fracdtdx$, but it involves $t$, which may seem objectionable to you (or your teacher). Fortunately, you started out with a formula that expresses $t$ in terms of $x$, and can plug that in to get
beginalign
fracdtdx &=frac(x-a)left(frac3vright)^2t-fracxv + frac1v\
&=frac(x-a)left(frac3vright)^2left(fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3) right)-fracxv + frac1v\
&=frac(x-a)left(frac3vright)^2left(fracsqrt(fraca2)^2+(a-x)^2(fracv3) right) + frac1v\
endalign
and there's your answer without using anything other than the chain rule and the (integer) power rule and a bit of algebra. It can, of course, be simplified via a bit more algebra, but I leave that to you.
answered Mar 24 at 17:02
John HughesJohn Hughes
65.4k24293
$endgroup$
add a comment |
$begingroup$
Hint: Treat $a$ and $v$ as constants. To differentiate this, you will need to use the chain rule. To start, differentiate $frac x v$ to get $frac 1 v$, and then note that $frac 3 v sqrtleft(frac a 2right)^2+(a-x)^2=frac 3 v fbig(g(x)big)$, where $g(x)=left(frac a 2right)^2+(a-x)^2$ and $f(u)=sqrt u$. Then apply the chain rule.
answered Mar 24 at 16:45
csch2csch2
6391314
$endgroup$
add a comment |
$begingroup$
Addition plus square roots plus chain rule.
The chain rule tells how to
differentiate a function of a function:
$(f(g(x)))' = g'(x) f'(g(x))
$
Example:
Since
$(x^2)' = 2x$
and
$(sqrtx)' = dfrac12sqrtx
$,
so
$(f(x)^2)' = 2f'(x)f(x)$
and
$(sqrtf(x))'
=dfracf'(x)2sqrtf(x)
$,
$beginarray\
(sqrt(fraca2)^2+(a-x)^2)'
&=((fraca2)^2+(a-x)^2)'dfrac12sqrt(fraca2)^2+(a-x)^2\
&=dfrac2(a-x)'(a-x)2sqrt(fraca2)^2+(a-x)^2\
&=dfrac-2(a-x)2sqrt(fraca2)^2+(a-x)^2\
endarray$
answered Mar 24 at 16:45
marty cohenmarty cohen
75.4k549130
$endgroup$
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
$begingroup$
Here's a sneaky way around some of the work others have suggested. In particular, you don't need to know how to differentiate the square-root function.
Do a little rewriting via algebra:
beginalign
t&=fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3)\
t-fracxv &=fracsqrt(fraca2)^2+(a-x)^2(fracv3)\
fracv3left(t-fracxvright) &=sqrt(fraca2)^2+(a-x)^2\
left(fracv3right)^2left(t-fracxvright)^2 &=(fraca2)^2+(a-x)^2
endalign
Now remembering that $t$ is really $t(x)$, a function that varies with $x$, we have a equality between two functions -- one on the left, one on the right. So their derivatives must be equal. The one on the left needs the power rule and the chain rule; the one on the right needs only the power rule. Differentiating, we get
beginalign
2left(fracv3right)^2left(t-fracxvright)left(fracdtdx + frac-1vright) &=2(a-x)(-1)\
left(fracv3right)^2left(t-fracxvright)left(fracdtdx + frac-1vright) &=(x-a)\
left(t-fracxvright)left(fracdtdx + frac-1vright) &=(x-a)left(frac3vright)^2\
fracdtdx + frac-1v &=frac(x-a)left(frac3vright)^2t-fracxv\
fracdtdx &=frac(x-a)left(frac3vright)^2t-fracxv + frac1v\
endalign
At this point, you have a formula for $fracdtdx$, but it involves $t$, which may seem objectionable to you (or your teacher). Fortunately, you started out with a formula that expresses $t$ in terms of $x$, and can plug that in to get
beginalign
fracdtdx &=frac(x-a)left(frac3vright)^2t-fracxv + frac1v\
&=frac(x-a)left(frac3vright)^2left(fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3) right)-fracxv + frac1v\
&=frac(x-a)left(frac3vright)^2left(fracsqrt(fraca2)^2+(a-x)^2(fracv3) right) + frac1v\
endalign
and there's your answer without using anything other than the chain rule and the (integer) power rule and a bit of algebra. It can, of course, be simplified via a bit more algebra, but I leave that to you.
answered Mar 24 at 17:02
John HughesJohn Hughes
65.4k24293
$endgroup$
add a comment |
$begingroup$
Here's a sneaky way around some of the work others have suggested. In particular, you don't need to know how to differentiate the square-root function.
Do a little rewriting via algebra:
beginalign
t&=fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3)\
t-fracxv &=fracsqrt(fraca2)^2+(a-x)^2(fracv3)\
fracv3left(t-fracxvright) &=sqrt(fraca2)^2+(a-x)^2\
left(fracv3right)^2left(t-fracxvright)^2 &=(fraca2)^2+(a-x)^2
endalign
Now remembering that $t$ is really $t(x)$, a function that varies with $x$, we have a equality between two functions -- one on the left, one on the right. So their derivatives must be equal. The one on the left needs the power rule and the chain rule; the one on the right needs only the power rule. Differentiating, we get
beginalign
2left(fracv3right)^2left(t-fracxvright)left(fracdtdx + frac-1vright) &=2(a-x)(-1)\
left(fracv3right)^2left(t-fracxvright)left(fracdtdx + frac-1vright) &=(x-a)\
left(t-fracxvright)left(fracdtdx + frac-1vright) &=(x-a)left(frac3vright)^2\
fracdtdx + frac-1v &=frac(x-a)left(frac3vright)^2t-fracxv\
fracdtdx &=frac(x-a)left(frac3vright)^2t-fracxv + frac1v\
endalign
At this point, you have a formula for $fracdtdx$, but it involves $t$, which may seem objectionable to you (or your teacher). Fortunately, you started out with a formula that expresses $t$ in terms of $x$, and can plug that in to get
beginalign
fracdtdx &=frac(x-a)left(frac3vright)^2t-fracxv + frac1v\
&=frac(x-a)left(frac3vright)^2left(fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3) right)-fracxv + frac1v\
&=frac(x-a)left(frac3vright)^2left(fracsqrt(fraca2)^2+(a-x)^2(fracv3) right) + frac1v\
endalign
and there's your answer without using anything other than the chain rule and the (integer) power rule and a bit of algebra. It can, of course, be simplified via a bit more algebra, but I leave that to you.
answered Mar 24 at 17:02
John HughesJohn Hughes
65.4k24293
$endgroup$
add a comment |
$begingroup$
Here's a sneaky way around some of the work others have suggested. In particular, you don't need to know how to differentiate the square-root function.
Do a little rewriting via algebra:
beginalign
t&=fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3)\
t-fracxv &=fracsqrt(fraca2)^2+(a-x)^2(fracv3)\
fracv3left(t-fracxvright) &=sqrt(fraca2)^2+(a-x)^2\
left(fracv3right)^2left(t-fracxvright)^2 &=(fraca2)^2+(a-x)^2
endalign
Now remembering that $t$ is really $t(x)$, a function that varies with $x$, we have a equality between two functions -- one on the left, one on the right. So their derivatives must be equal. The one on the left needs the power rule and the chain rule; the one on the right needs only the power rule. Differentiating, we get
beginalign
2left(fracv3right)^2left(t-fracxvright)left(fracdtdx + frac-1vright) &=2(a-x)(-1)\
left(fracv3right)^2left(t-fracxvright)left(fracdtdx + frac-1vright) &=(x-a)\
left(t-fracxvright)left(fracdtdx + frac-1vright) &=(x-a)left(frac3vright)^2\
fracdtdx + frac-1v &=frac(x-a)left(frac3vright)^2t-fracxv\
fracdtdx &=frac(x-a)left(frac3vright)^2t-fracxv + frac1v\
endalign
At this point, you have a formula for $fracdtdx$, but it involves $t$, which may seem objectionable to you (or your teacher). Fortunately, you started out with a formula that expresses $t$ in terms of $x$, and can plug that in to get
beginalign
fracdtdx &=frac(x-a)left(frac3vright)^2t-fracxv + frac1v\
&=frac(x-a)left(frac3vright)^2left(fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3) right)-fracxv + frac1v\
&=frac(x-a)left(frac3vright)^2left(fracsqrt(fraca2)^2+(a-x)^2(fracv3) right) + frac1v\
endalign
and there's your answer without using anything other than the chain rule and the (integer) power rule and a bit of algebra. It can, of course, be simplified via a bit more algebra, but I leave that to you.
answered Mar 24 at 17:02
John HughesJohn Hughes
65.4k24293
$endgroup$
Here's a sneaky way around some of the work others have suggested. In particular, you don't need to know how to differentiate the square-root function.
Do a little rewriting via algebra:
beginalign
t&=fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3)\
t-fracxv &=fracsqrt(fraca2)^2+(a-x)^2(fracv3)\
fracv3left(t-fracxvright) &=sqrt(fraca2)^2+(a-x)^2\
left(fracv3right)^2left(t-fracxvright)^2 &=(fraca2)^2+(a-x)^2
endalign
Now remembering that $t$ is really $t(x)$, a function that varies with $x$, we have a equality between two functions -- one on the left, one on the right. So their derivatives must be equal. The one on the left needs the power rule and the chain rule; the one on the right needs only the power rule. Differentiating, we get
beginalign
2left(fracv3right)^2left(t-fracxvright)left(fracdtdx + frac-1vright) &=2(a-x)(-1)\
left(fracv3right)^2left(t-fracxvright)left(fracdtdx + frac-1vright) &=(x-a)\
left(t-fracxvright)left(fracdtdx + frac-1vright) &=(x-a)left(frac3vright)^2\
fracdtdx + frac-1v &=frac(x-a)left(frac3vright)^2t-fracxv\
fracdtdx &=frac(x-a)left(frac3vright)^2t-fracxv + frac1v\
endalign
At this point, you have a formula for $fracdtdx$, but it involves $t$, which may seem objectionable to you (or your teacher). Fortunately, you started out with a formula that expresses $t$ in terms of $x$, and can plug that in to get
beginalign
fracdtdx &=frac(x-a)left(frac3vright)^2t-fracxv + frac1v\
&=frac(x-a)left(frac3vright)^2left(fracxv+fracsqrt(fraca2)^2+(a-x)^2(fracv3) right)-fracxv + frac1v\
&=frac(x-a)left(frac3vright)^2left(fracsqrt(fraca2)^2+(a-x)^2(fracv3) right) + frac1v\
endalign
and there's your answer without using anything other than the chain rule and the (integer) power rule and a bit of algebra. It can, of course, be simplified via a bit more algebra, but I leave that to you.
answered Mar 24 at 17:02
John HughesJohn Hughes
65.4k24293
edited Mar 24 at 17:21
answered Mar 24 at 17:02
John HughesJohn Hughes
65.4k24293
answered Mar 24 at 17:02
John HughesJohn Hughes
65.4k24293
answered Mar 24 at 17:02
John HughesJohn Hughes
65.4k24293
65.4k24293
add a comment |
add a comment |
$begingroup$
Hint: Treat $a$ and $v$ as constants. To differentiate this, you will need to use the chain rule. To start, differentiate $frac x v$ to get $frac 1 v$, and then note that $frac 3 v sqrtleft(frac a 2right)^2+(a-x)^2=frac 3 v fbig(g(x)big)$, where $g(x)=left(frac a 2right)^2+(a-x)^2$ and $f(u)=sqrt u$. Then apply the chain rule.
answered Mar 24 at 16:45
csch2csch2
6391314
$endgroup$
add a comment |
$begingroup$
Hint: Treat $a$ and $v$ as constants. To differentiate this, you will need to use the chain rule. To start, differentiate $frac x v$ to get $frac 1 v$, and then note that $frac 3 v sqrtleft(frac a 2right)^2+(a-x)^2=frac 3 v fbig(g(x)big)$, where $g(x)=left(frac a 2right)^2+(a-x)^2$ and $f(u)=sqrt u$. Then apply the chain rule.
answered Mar 24 at 16:45
csch2csch2
6391314
$endgroup$
add a comment |
$begingroup$
Hint: Treat $a$ and $v$ as constants. To differentiate this, you will need to use the chain rule. To start, differentiate $frac x v$ to get $frac 1 v$, and then note that $frac 3 v sqrtleft(frac a 2right)^2+(a-x)^2=frac 3 v fbig(g(x)big)$, where $g(x)=left(frac a 2right)^2+(a-x)^2$ and $f(u)=sqrt u$. Then apply the chain rule.
answered Mar 24 at 16:45
csch2csch2
6391314
$endgroup$
Hint: Treat $a$ and $v$ as constants. To differentiate this, you will need to use the chain rule. To start, differentiate $frac x v$ to get $frac 1 v$, and then note that $frac 3 v sqrtleft(frac a 2right)^2+(a-x)^2=frac 3 v fbig(g(x)big)$, where $g(x)=left(frac a 2right)^2+(a-x)^2$ and $f(u)=sqrt u$. Then apply the chain rule.
answered Mar 24 at 16:45
csch2csch2
6391314
answered Mar 24 at 16:45
csch2csch2
6391314
answered Mar 24 at 16:45
csch2csch2
6391314
answered Mar 24 at 16:45
csch2csch2
6391314
6391314
add a comment |
add a comment |
$begingroup$
Addition plus square roots plus chain rule.
The chain rule tells how to
differentiate a function of a function:
$(f(g(x)))' = g'(x) f'(g(x))
$
Example:
Since
$(x^2)' = 2x$
and
$(sqrtx)' = dfrac12sqrtx
$,
so
$(f(x)^2)' = 2f'(x)f(x)$
and
$(sqrtf(x))'
=dfracf'(x)2sqrtf(x)
$,
$beginarray\
(sqrt(fraca2)^2+(a-x)^2)'
&=((fraca2)^2+(a-x)^2)'dfrac12sqrt(fraca2)^2+(a-x)^2\
&=dfrac2(a-x)'(a-x)2sqrt(fraca2)^2+(a-x)^2\
&=dfrac-2(a-x)2sqrt(fraca2)^2+(a-x)^2\
endarray$
answered Mar 24 at 16:45
marty cohenmarty cohen
75.4k549130
$endgroup$
add a comment |
$begingroup$
Addition plus square roots plus chain rule.
The chain rule tells how to
differentiate a function of a function:
$(f(g(x)))' = g'(x) f'(g(x))
$
Example:
Since
$(x^2)' = 2x$
and
$(sqrtx)' = dfrac12sqrtx
$,
so
$(f(x)^2)' = 2f'(x)f(x)$
and
$(sqrtf(x))'
=dfracf'(x)2sqrtf(x)
$,
$beginarray\
(sqrt(fraca2)^2+(a-x)^2)'
&=((fraca2)^2+(a-x)^2)'dfrac12sqrt(fraca2)^2+(a-x)^2\
&=dfrac2(a-x)'(a-x)2sqrt(fraca2)^2+(a-x)^2\
&=dfrac-2(a-x)2sqrt(fraca2)^2+(a-x)^2\
endarray$
answered Mar 24 at 16:45
marty cohenmarty cohen
75.4k549130
$endgroup$
add a comment |
$begingroup$
Addition plus square roots plus chain rule.
The chain rule tells how to
differentiate a function of a function:
$(f(g(x)))' = g'(x) f'(g(x))
$
Example:
Since
$(x^2)' = 2x$
and
$(sqrtx)' = dfrac12sqrtx
$,
so
$(f(x)^2)' = 2f'(x)f(x)$
and
$(sqrtf(x))'
=dfracf'(x)2sqrtf(x)
$,
$beginarray\
(sqrt(fraca2)^2+(a-x)^2)'
&=((fraca2)^2+(a-x)^2)'dfrac12sqrt(fraca2)^2+(a-x)^2\
&=dfrac2(a-x)'(a-x)2sqrt(fraca2)^2+(a-x)^2\
&=dfrac-2(a-x)2sqrt(fraca2)^2+(a-x)^2\
endarray$
answered Mar 24 at 16:45
marty cohenmarty cohen
75.4k549130
$endgroup$
Addition plus square roots plus chain rule.
The chain rule tells how to
differentiate a function of a function:
$(f(g(x)))' = g'(x) f'(g(x))
$
Example:
Since
$(x^2)' = 2x$
and
$(sqrtx)' = dfrac12sqrtx
$,
so
$(f(x)^2)' = 2f'(x)f(x)$
and
$(sqrtf(x))'
=dfracf'(x)2sqrtf(x)
$,
$beginarray\
(sqrt(fraca2)^2+(a-x)^2)'
&=((fraca2)^2+(a-x)^2)'dfrac12sqrt(fraca2)^2+(a-x)^2\
&=dfrac2(a-x)'(a-x)2sqrt(fraca2)^2+(a-x)^2\
&=dfrac-2(a-x)2sqrt(fraca2)^2+(a-x)^2\
endarray$
answered Mar 24 at 16:45
marty cohenmarty cohen
75.4k549130
answered Mar 24 at 16:45
marty cohenmarty cohen
75.4k549130
answered Mar 24 at 16:45
marty cohenmarty cohen
75.4k549130
answered Mar 24 at 16:45
marty cohenmarty cohen
75.4k549130
75.4k549130
add a comment |
add a comment |
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