How do you find the slope of the curve #y^3-xy^2=4# at the point where y=2?

Answer 1

The slope of the curve is #1/2#

We need to start by finding the x-coordinate.

#2^3 -x(2)^2= 4#
#-4x = -4#
#x=1#

Now we differentiate.

#3y^2(dy/dx) -y^2 - 2xy(dy/dx) =0#
#dy/dx(3y^2 -2xy) = y^2#
#dy/dx = y/(3y - 2x)#
The slope is simply the value of #dy/dx# at #(1,2)#
#dy/dx = 2/(6- 2) = 2/4=1/2#

Hopefully this helps!

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Answer 2

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This yields the slope of the curve at the point where ( y = 2To find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the pointTo find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 \To find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. 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After differentiationTo find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y =To find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation andTo find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. 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This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution,To find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 \To find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, theTo find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 ) isTo find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, the slope isTo find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 ) is (To find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, the slope is foundTo find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 ) is ( \To find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, the slope is found toTo find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 ) is ( \fracTo find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, the slope is found to beTo find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 ) is ( \frac{-To find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, the slope is found to be (To find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 ) is ( \frac{-3To find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, the slope is found to be ( \To find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 ) is ( \frac{-3}{To find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, the slope is found to be ( \fracTo find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 ) is ( \frac{-3}{4To find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, the slope is found to be ( \frac{{To find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 ) is ( \frac{-3}{4}To find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, the slope is found to be ( \frac{{dyTo find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 ) is ( \frac{-3}{4} \To find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, the slope is found to be ( \frac{{dy}}{{To find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 ) is ( \frac{-3}{4} ).To find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, the slope is found to be ( \frac{{dy}}{{dxTo find the slope of the curve ( y^3 - xy^2 = 4 ) at the point where ( y = 2 ), first, differentiate the equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for the slope. After differentiation and substitution, the slope at the point where ( y = 2 ) is ( \frac{-3}{4} ).To find the slope of the curve at the point where ( y = 2 ), first differentiate the given equation implicitly with respect to ( x ). Then, substitute ( y = 2 ) into the resulting expression and solve for ( \frac{{dy}}{{dx}} ). This yields the slope of the curve at the point where ( y = 2 ). After differentiation and substitution, the slope is found to be ( \frac{{dy}}{{dx}} = \frac{{12x - 4}}{{3y^2 - 2xy}} ). Substituting ( y = 2 ), the slope at the point where ( y = 2 ) is ( \frac{{12x - 4}}{{12}} = x - \frac{1}{3} ).

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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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