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Oksi-84 [34.3K]
3 years ago
7

Which company is better

Mathematics
1 answer:
lana [24]3 years ago
8 0

Answer:

both because thats what i think

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Find dy/dx of the function y = √x sec*-1 (√x)​
ELEN [110]

Hi there!

\large\boxed{\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) +  \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}}

y = \sqrt{x} * sec^{-1}(-\sqrt{x}})

Use the chain rule and multiplication rules to solve:

g(x) * f(x) = f'(x)g(x) + g'(x)f(x)

g(f(x)) = g'(f(x)) * 'f(x))

Thus:

f(x) = √x

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\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{\sqrt{x}\sqrt{\sqrt{x}^{2} - 1}} * \frac{1}{2\sqrt{x}}

Simplify:

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) + \sqrt{x} * \frac{1}{2|x|\sqrt{{x} - 1}}

\frac{dy}{dx} = \frac{1}{2\sqrt{x}}sec^{-1}(\sqrt{x}) +  \frac{1}{2|\sqrt{x}|\sqrt{{x} - 1}}

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