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junior college 1 | H1 Maths
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Need help for 5(iii) and 6 urgent thanks
Date Posted: 3 years ago
Views: 419
dS/dt = 150
dx/dt = dS/dt × dx/dS
= dS/dt ÷ dS/dx
We know from the first part that :
① h = 300/x²
② dS/dx = 24x - 3000/x²
When h = 30,
30 = 300/x²
x² = 300/30
x² = 10
x = √10
(since length is positive, x must be positive)
So dS/dx = 24√10 - 3000/10
= 24√10 - 300
Therefore, dx/dt
= 150 / (24√10 - 300)
= 25 / (4√10 - 50)
= 25 x (4√10 + 50) / (4√10 + 50)(4√10 - 50)
= (100√10 + 1250) / (160 - 2500)
= (100√10 + 1250) / (-2340)
= - (10√10 + 125) / 234
dx/dt = dS/dt × dx/dS
= dS/dt ÷ dS/dx
We know from the first part that :
① h = 300/x²
② dS/dx = 24x - 3000/x²
When h = 30,
30 = 300/x²
x² = 300/30
x² = 10
x = √10
(since length is positive, x must be positive)
So dS/dx = 24√10 - 3000/10
= 24√10 - 300
Therefore, dx/dt
= 150 / (24√10 - 300)
= 25 / (4√10 - 50)
= 25 x (4√10 + 50) / (4√10 + 50)(4√10 - 50)
= (100√10 + 1250) / (160 - 2500)
= (100√10 + 1250) / (-2340)
= - (10√10 + 125) / 234
Edited
d/dx ln (x^e / (1 + x) )
= (d/dx (x^e / (1+x)) ) / (x^e / (1 + x) )
= [ ( (1+x)(ex^(e-1)) - x^e ) / (1 + x)² ] / (x^e / (1+x) )
= [ (1+x)(ex^e)/x - x^e ) / (1 + x)² ] × (1+x) / x^e
= [(1+x)(e/x) - 1] / (1+x)
= ((1+x)e - x) / x(1+x)
= (e + xe - x) / x(1+x)
= (e + x(e - 1)) / x(1+x)
(Shown)
= (d/dx (x^e / (1+x)) ) / (x^e / (1 + x) )
= [ ( (1+x)(ex^(e-1)) - x^e ) / (1 + x)² ] / (x^e / (1+x) )
= [ (1+x)(ex^e)/x - x^e ) / (1 + x)² ] × (1+x) / x^e
= [(1+x)(e/x) - 1] / (1+x)
= ((1+x)e - x) / x(1+x)
= (e + xe - x) / x(1+x)
= (e + x(e - 1)) / x(1+x)
(Shown)
Edited. This are the steps that uses quotient rule. The rest of it is just cancelling common terms.
Thank you so much for the explanation!!
Welcome.
See 2 Answers
Do we need to differentiate ‘e’ when d/dx[ e ln x]
e is a constant (≈ 2.71828183) so no.
But isn’t it inside the bracket? And if I differentiate a constant(e) it will become zero right ?
We can also consider the product rule where differentiating e ln x will give us
e d/dx ln x PLUS ln x times d/dx e
First term is e/x
Second term after the plus is ln x times 0, which is the differentiating constant numbers thing.
e d/dx ln x PLUS ln x times d/dx e
First term is e/x
Second term after the plus is ln x times 0, which is the differentiating constant numbers thing.
Constants can always be taken out.
Eg. Lets say y = 2x
dy/dx = d/dx (2x) = 2
But we can also say
dy/dx = d/dx (2x)
= 2 d/dx (x)
= 2 (1)
= 2
The result is the same.
Eg. Lets say y = 2x
dy/dx = d/dx (2x) = 2
But we can also say
dy/dx = d/dx (2x)
= 2 d/dx (x)
= 2 (1)
= 2
The result is the same.
To verify that the constant can be taken out :
Use product rule like Eric mentioned.
① d/dx ( e lnx)
= e (d/dx (lnx)) + (d/dx (e)) lnx
= e (1/x) + (0)lnx
= e (1/x) + 0
= e/x
② d/dx (e lnx)
= e (d/dx (lnx))
= e (1/x)
= e/x
We get the same result for ① and ②
Perhaps the e is confusing you. Just note that it's a constant.
Eg. Let's say I replace e with 2.718,
d/dx (2.718 lnx)
= 2.71(1/x)
= 2.71/x
Likewise,
if I replace e with another constant like π,
Then d/dx (π lnx)
= π (1/x)
= π/x
Use product rule like Eric mentioned.
① d/dx ( e lnx)
= e (d/dx (lnx)) + (d/dx (e)) lnx
= e (1/x) + (0)lnx
= e (1/x) + 0
= e/x
② d/dx (e lnx)
= e (d/dx (lnx))
= e (1/x)
= e/x
We get the same result for ① and ②
Perhaps the e is confusing you. Just note that it's a constant.
Eg. Let's say I replace e with 2.718,
d/dx (2.718 lnx)
= 2.71(1/x)
= 2.71/x
Likewise,
if I replace e with another constant like π,
Then d/dx (π lnx)
= π (1/x)
= π/x
Once we know the constant can be taken out, we can save all the trouble of differentiating it via product rule
done
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clear
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For Q5iii we need to obtain dS/dx when h = 30 before finding dx/dt. Of course, our expression for dS/dx is in terms of x and not h, so we need to find the equivalent value of x for the given value of h.
Date Posted:
3 years ago
Thank you!!!
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