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3 years ago

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