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3 years ago

Determine whether the relation is a function. quick EMERGENCY I NEED HELP

Mathematics

1 answer:

saul85 [17]3 years ago

6 0

Not a function.
It goes from -1=-4 to -1=-2

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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:

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

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3 years ago

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3 years ago

Determine whether the relation is a function. quick EMERGENCY I NEED HELP ​

Determine whether the relation is a function. quick EMERGENCY I NEED HELP