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How do you implicitly differentiate #-3=xsecy#?

Answer 1

#(dy)/(dx)=-xcoty#

When we implicitly differentiate a function #f(y)#, we first differentiate #f(y)# with respect to #y# and then multiply by #(dy)/(dx)#. This comes from chain rule as #(df)/(dx)=(df)/(dy)xx(dy)/(dx)#.

Now using product rule for differentiating #3=xsecy#, we get

#0=x xxsecy+1xxsecytany(dy)/(dx)# or

#secytany(dy)/(dx)=-xsecy# or

#(dy)/(dx)=(-xsecy)/(secytany)#

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

= #-xcoty#

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Answer 2

Samantha Adkison

To implicitly differentiate -3 = xsec(y) with respect to x, follow these steps:

Differentiate both sides of the equation with respect to x.
Apply the chain rule where necessary.
Solve for dy/dx, which represents the derivative of y with respect to x.

Here are the steps:

-3 = xsec(y)

Differentiate both sides with respect to x:

0 = d/dx(xsec(y))

Apply the chain rule:

0 = sec(y) * (d(x)/dx) + x * d(sec(y))/dx

Using the derivatives of sec(y) and sec(y)tan(y):

0 = sec(y) * (1) + x * (sec(y)tan(y) * (dy/dx))

Rearrange the equation to solve for dy/dx:

dy/dx = -sec(y)/x * tan(y)

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.