How do you implicitly differentiate $- 3 = x sec y$ $-3=xsecy$ ?

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How do you implicitly differentiate $- 3 = x sec y$ $-3=xsecy$ ?

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Answer 1 · 2016-07-21T08:33:30

$\frac{d y}{d x} = - x cot y$ $(dy)/(dx)=-xcoty$

Explanation:

When we implicitly differentiate a function $f (y)$ $f(y)$ , we first differentiate $f (y)$ $f(y)$ with respect to $y$ $y$ and then multiply by $\frac{d y}{d x}$ $(dy)/(dx)$ . This comes from chain rule as $\frac{d f}{d x} = \frac{d f}{d y} \times \frac{d y}{d x}$ $(df)/(dx)=(df)/(dy)xx(dy)/(dx)$ .

Now using product rule for differentiating $3 = x sec y$ $3=xsecy$ , we get

$0 = x \times sec y + 1 \times sec y tan y \frac{d y}{d x}$ $0=x xxsecy+1xxsecytany(dy)/(dx)$ or

$sec y tan y \frac{d y}{d x} = - x sec y$ $secytany(dy)/(dx)=-xsecy$ or

$\frac{d y}{d x} = \frac{- x sec y}{sec y tan y}$ $(dy)/(dx)=(-xsecy)/(secytany)$

= $(-xcancel(secy))/(cancel(secy)tany)$

= $-xcoty$

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How do you implicitly differentiate $- 3 = x sec y$ $-3=xsecy$ ?

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