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GaryK [48]
3 years ago
14

PLEASE HELP ME WITH THE SECOND ONE ;(

Mathematics
1 answer:
tensa zangetsu [6.8K]3 years ago
8 0
Im pretty sure it’s joint
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2xy=4(1/2xy) because
frutty [35]
Because, when 4 multiply by 1/2 will equal 2, so 2xy=4(1/2xy)
8 0
3 years ago
Y=x.arctan(x)^1/2. find dy/dx. pls show steps​
Whitepunk [10]

y=x(\arctan x)^{1/2}

Use the product rule first:

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{\mathrm dx}{\mathrm dx}(\arctan x)^{1/2}+x\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}

\dfrac{\mathrm dy}{\mathrm dx}=(\arctan x)^{1/2}+x\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}

Use the chain rule to compute the derivative of (\arctan x)^{1/2}. Let z=(\arctan x)^{1/2} and take u=\arctan x, so that by the chain rule

\dfrac{\mathrm dz}{\mathrm dx}=\dfrac{\mathrm dz}{\mathrm du}\dfrac{\mathrm du}{\mathrm dx}

\dfrac{\mathrm dz}{\mathrm du}=\dfrac{\mathrm du^{1/2}}{\mathrm du}=\dfrac12u^{-1/2}

\dfrac{\mathrm du}{\mathrm dx}=\dfrac{\mathrm d\arctan x}{\mathrm dx}=\dfrac1{1+x^2}

\implies\dfrac{\mathrm d(\arctan x)^{1/2}}{\mathrm dx}=\dfrac1{2(\arctan x)^{1/2}(1+x^2)}

So we have

\dfrac{\mathrm dy}{\mathrm dx}=(\arctan x)^{1/2}+\dfrac x{2(\arctan x)^{1/2}(1+x^2)}

You can rewrite this a bit by factoring (\arctan x)^{-1/2}, just to make it look neater:

\dfrac{\mathrm dy}{\mathrm dx}=\dfrac1{2(\arctan x)^{1/2}}\left(2\arctan x+\dfrac x{1+x^2}\right)

3 0
3 years ago
If x = -2, which inequality is true?
Anettt [7]
It’s J, 3 - X would equal 3 - (-2) which turns into 3+2 which equals 5 but 5 is not greater than 10
4 0
3 years ago
Read 2 more answers
All angles in the figure at the right are right angles what is the agree of the figure
wel

Answer:

Ans: 36

Step-by-step explanation:

just separate it into three rectangles and use the formula

A=l*b

and at last add all of them

6 0
3 years ago
Domain of f(x)=(1/4)^x
ANEK [815]

Answer: B

Step-by-step explanation:

The domain of a function is the set of x-values.

6 0
2 years ago
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