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Cloud [144]
2 years ago
12

Peyton buys cookies and apples at the store.

Mathematics
1 answer:
Dmitrij [34]2 years ago
6 0

Answer:

Hope this helps sorry I'm in class right now

Step-by-step explanation:

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Susan's stylist has done a great job of coloring her hair. She wants to leave a 20 percent tip. If the bill was $150, how much s
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She should leave $30. Hope it helps! If you could vote me brainiest, that would be awesome!
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Find dy/dx if f(x) = (x + 8)^3x.
vlabodo [156]
This is quite a complex problem. I wrote out a really nice solution but I can't work out how to put it on the website as the app is very poorly made. Still, I'll just have to type it all in...

Okay so you need to use a technique called logarithmic differentiation. It seems quite unnatural to start with but the result is very impressive.

Let y = (x+8)^(3x)

Take the natural log of both sides:

ln(y) = ln((x+8)^(3x))

By laws of logarithms, this can be rearranged:

ln(y) = 3xln(x+8)

Next, differentiate both sides. By implicit differentiation:

d/dx(ln(y)) = 1/y dy/dx

The right hand side is harder to differentiate. Using the substitution u = 3x and v = ln(x+8):

d/dx(3xln(x+8)) = d/dx(uv)

du/dx = 3

Finding dv/dx is harder, and involves the chain rule. Let a = x+ 8:

v = ln(a)
da/dx = 1
dv/da = 1/a

By chain rule:

dv/dx = dv/da * da/dx = 1/a = 1/(x+8)

Finally, use the product rule:

d/dx(uv) = u * dv/dx + v * du/dx = 3x/(x+8) + 3ln(x+8)

This overall produces the equation:

1/y * dy/dx = 3x/(x+8) + 3ln(x+8)

We want to solve for dy/dx, achievable by multiplying both sides by y:

dy/dx = y(3x/(x+8) + 3ln(x+8))

Since we know y = (x+8)^(3x):

dy/dx = ((x+8)^(3x))(3x/(x+8) + 3ln(x+8))

Neatening this up a bit, we factorise out 3/(x+8):

dy/dx = (3(x+8)^(3x-1))(x + (x+8)ln(x+8))

Well wasn't that a marathon? It's a nightmare typing that in, I hope you can follow all the steps.

I hope this helped you :)
8 0
3 years ago
Which of these number is an integer? A. −4.0 B. 1.50 C. -5/8 D. infinity
NNADVOKAT [17]
<span>−4.0 is an integer.

hope it helps</span>
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