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sesenic [268]
3 years ago
6

Find dy dx of y equals the secant of the square root of x.

Mathematics
1 answer:
iVinArrow [24]3 years ago
5 0

\bf y=sec(\sqrt{x})\implies y=sec\left( x^{\frac{1}{2}} \right)\implies \cfrac{dy}{dx}=\stackrel{\textit{chain rule}}{sec\left( x^{\frac{1}{2}} \right)tan\left( x^{\frac{1}{2}} \right)\cdot \cfrac{1}{2}x^{-\frac{1}{2}}} \\\\\\ \cfrac{dy}{dx}=sec\left( x^{\frac{1}{2}} \right)tan\left( x^{\frac{1}{2}} \right)\cdot \cfrac{1}{2x^{\frac{1}{2}}}\implies \cfrac{dy}{dx}=\cfrac{sec\left( \sqrt{x} \right)tan\left( \sqrt{x} \right)}{2\sqrt{x}}

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Answer:

Step-by-step explanation:

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The second part zeros of 1 means (x-1) and multiplicity 1 means (x-1).

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3 years ago
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So in order to solve you must first combine like terms.
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3 years ago
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