How do you implicitly differentiate #-1=xytan(x/y) #?

Answer 1

#frac{dy}{dx} = frac{y}{x}*frac{ 2xcsc(frac{2x}{y}) + y }{ xcot(x/y) - y }#

#-1 = xytan(x/y)#
Differentiate both sides w.r.t. #x#.
#frac{d}{dx}(-1) = frac{d}{dx}(xytan(x/y))#
#0 = frac{d}{dx}(xytan(x/y))#

Use the Product Rule, Chain Rule, and Quotient Rule.

#frac{d}{dx}(xytan(x/y)) = xyfrac{d}{dx}(tan(x/y)) + tan(x/y)frac{d}{dx}(xy)#
#= xyfrac{d}{dx}(tan(x/y)) + tan(x/y)(yfrac{d}{dx}(x) + xfrac{d}{dx}(y))#
#= xysec^2(x/y)frac{d}{dx}(x/y) + tan(x/y)(y + xfrac{dy}{dx})#
#= xysec^2(x/y)frac{yfrac{d}{dx}(x) - xfrac{d}{dx}(y)}{y^2} + tan(x/y)(y +xfrac{dy}{dx})#
#= frac{x}{y}sec^2(x/y)(y - xfrac{dy}{dx}) + tan(x/y)(y + xfrac{dy}{dx})#
#= xsec^2(x/y) - frac{x^2}{y}frac{dy}{dx} + ytan(x/y)+xtan(x/y)frac{dy}{dx}#
#= xsec^2(x/y) + ytan(x/y) + (xtan(x/y) - frac{x^2}{y})frac{dy}{dx} = 0#
#frac{dy}{dx} = - frac{ xsec^2(x/y) + ytan(x/y) }{ xtan(x/y) - frac{x^2}{y} }#
#= frac{y}{x}*frac{ 2xcsc(frac{2x}{y}) + y }{ xcot(x/y) - y }#
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Answer 2

To implicitly differentiate the equation ( -1 = xy \tan(\frac{x}{y}) ), follow these steps:

  1. Differentiate both sides of the equation with respect to ( x ).
  2. Use the product rule for the term ( xy ).
  3. Use the chain rule for the term ( \tan(\frac{x}{y}) ).
  4. Solve for ( \frac{dy}{dx} ).

Differentiating ( -1 = xy \tan(\frac{x}{y}) ) with respect to ( x ):

[ \frac{d}{dx}(-1) = \frac{d}{dx}(xy \tan(\frac{x}{y})) ]

Using the product rule for ( xy ):

[ 0 = y\frac{dx}{dx} + x\frac{dy}{dx} + xy \frac{d}{dx}(\tan(\frac{x}{y})) ]

Simplify and apply the chain rule for ( \tan(\frac{x}{y}) ):

[ 0 = y + x\frac{dy}{dx} + xy \left( \sec^2(\frac{x}{y}) \frac{d}{dx}(\frac{x}{y}) \right) ]

Differentiating ( \frac{x}{y} ) with respect to ( x ):

[ \frac{d}{dx}(\frac{x}{y}) = \frac{1}{y} - \frac{x}{y^2} \frac{dy}{dx} ]

Substitute back into the equation:

[ 0 = y + x\frac{dy}{dx} + xy \left( \sec^2(\frac{x}{y}) \left( \frac{1}{y} - \frac{x}{y^2} \frac{dy}{dx} \right) \right) ]

Simplify:

[ 0 = y + x\frac{dy}{dx} + \frac{x}{y} \sec^2(\frac{x}{y}) - x^2 \sec^2(\frac{x}{y}) \frac{dy}{dx} ]

Isolate ( \frac{dy}{dx} ):

[ x\frac{dy}{dx} - x^2 \sec^2(\frac{x}{y}) \frac{dy}{dx} = -y - \frac{x}{y} \sec^2(\frac{x}{y}) ]

[ \frac{dy}{dx}(x - x^2 \sec^2(\frac{x}{y})) = -y - \frac{x}{y} \sec^2(\frac{x}{y}) ]

[ \frac{dy}{dx} = \frac{-y - \frac{x}{y} \sec^2(\frac{x}{y})}{x - x^2 \sec^2(\frac{x}{y})} ]

This is the implicit derivative of ( y ) with respect to ( x ).

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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.

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.

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.

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.

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