Find y″ by differentiating implicitly twice, and evaluate it at a point on the curve. 56 problems with worked solutions; in 56 of them every
equation is proved by a computer algebra system.
Try each one before opening the solution. Every line is labeled with what checked it. How we verify
The curve \( \displaystyle x^{2} + y^{2} = 2 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle y'' \) there.
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\[ \left. x^{2} + y^{2} \right|_{\substack{ x=1\\ y=1 }} = 2 \]
The point is on the curve.✓ Proved
\[ - \frac{\frac{\partial}{\partial x} \left(x^{2} + y^{2}\right)}{\frac{\partial}{\partial y} \left(x^{2} + y^{2}\right)} = - \frac{x}{y} \]
y′ = −F_x/F_y.✓ Proved
\[ \frac{d}{d x} \left(- \frac{x}{Y{\left(x \right)}}\right) = \frac{x \frac{d}{d x} Y{\left(x \right)}}{Y^{2}{\left(x \right)}} - \frac{1}{Y{\left(x \right)}} \]
Differentiate y′ again, with y a function of x.✓ Proved
Replace dy/dx in that by y′ itself, and simplify.
Reviewed
\[ \left. \frac{- x^{2} - y^{2}}{y^{3}} \right|_{\substack{ x=1\\ y=1 }} = -2 \]
At (1, 1).✓ Proved
Answer \( y'' = \frac{- x^{2} - y^{2}}{y^{3}},\quad y''(1) = -2 \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The curve \( \displaystyle 3 x^{2} + 2 y^{2} = 5 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle y'' \) there.
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\[ \left. 3 x^{2} + 2 y^{2} \right|_{\substack{ x=1\\ y=1 }} = 5 \]
The point is on the curve.✓ Proved
\[ - \frac{\frac{\partial}{\partial x} \left(3 x^{2} + 2 y^{2}\right)}{\frac{\partial}{\partial y} \left(3 x^{2} + 2 y^{2}\right)} = - \frac{3 x}{2 y} \]
y′ = −F_x/F_y.✓ Proved
\[ \frac{d}{d x} \left(- \frac{3 x}{2 Y{\left(x \right)}}\right) = \frac{3 x \frac{d}{d x} Y{\left(x \right)}}{2 Y^{2}{\left(x \right)}} - \frac{3}{2 Y{\left(x \right)}} \]
Differentiate y′ again, with y a function of x.✓ Proved
Replace dy/dx in that by y′ itself, and simplify.
Reviewed
\[ \left. \frac{3 \left(- 3 x^{2} - 2 y^{2}\right)}{4 y^{3}} \right|_{\substack{ x=1\\ y=1 }} = - \frac{15}{4} \]
At (1, 1).✓ Proved
Answer \( y'' = \frac{3 \left(- 3 x^{2} - 2 y^{2}\right)}{4 y^{3}},\quad y''(1) = - \frac{15}{4} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The curve \( \displaystyle 4 x^{2} + y^{2} = 5 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle y'' \) there.
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\[ \left. 4 x^{2} + y^{2} \right|_{\substack{ x=1\\ y=-1 }} = 5 \]
The point is on the curve.✓ Proved
\[ - \frac{\frac{\partial}{\partial x} \left(4 x^{2} + y^{2}\right)}{\frac{\partial}{\partial y} \left(4 x^{2} + y^{2}\right)} = - \frac{4 x}{y} \]
y′ = −F_x/F_y.✓ Proved
\[ \frac{d}{d x} \left(- \frac{4 x}{Y{\left(x \right)}}\right) = \frac{4 x \frac{d}{d x} Y{\left(x \right)}}{Y^{2}{\left(x \right)}} - \frac{4}{Y{\left(x \right)}} \]
Differentiate y′ again, with y a function of x.✓ Proved
Replace dy/dx in that by y′ itself, and simplify.
Reviewed
\[ \left. - \frac{16 x^{2}}{y^{3}} - \frac{4}{y} \right|_{\substack{ x=1\\ y=-1 }} = 20 \]
At (1, -1).✓ Proved
Answer \( y'' = - \frac{16 x^{2}}{y^{3}} - \frac{4}{y},\quad y''(1) = 20 \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The curve \( \displaystyle 2 x^{2} + 2 y^{2} = 4 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle y'' \) there.
Show the first step
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\[ \left. 2 x^{2} + 2 y^{2} \right|_{\substack{ x=1\\ y=-1 }} = 4 \]
The point is on the curve.✓ Proved
\[ - \frac{\frac{\partial}{\partial x} \left(2 x^{2} + 2 y^{2}\right)}{\frac{\partial}{\partial y} \left(2 x^{2} + 2 y^{2}\right)} = - \frac{x}{y} \]
y′ = −F_x/F_y.✓ Proved
\[ \frac{d}{d x} \left(- \frac{x}{Y{\left(x \right)}}\right) = \frac{x \frac{d}{d x} Y{\left(x \right)}}{Y^{2}{\left(x \right)}} - \frac{1}{Y{\left(x \right)}} \]
Differentiate y′ again, with y a function of x.✓ Proved
Replace dy/dx in that by y′ itself, and simplify.
Reviewed
\[ \left. \frac{- x^{2} - y^{2}}{y^{3}} \right|_{\substack{ x=1\\ y=-1 }} = 2 \]
At (1, -1).✓ Proved
Answer \( y'' = \frac{- x^{2} - y^{2}}{y^{3}},\quad y''(1) = 2 \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The curve \( \displaystyle x^{2} + y^{2} = 25 \) passes through \( \displaystyle (-3, 4) \). Find \( \displaystyle y'' \) there.
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\[ \left. x^{2} + y^{2} \right|_{\substack{ x=-3\\ y=4 }} = 25 \]
The point is on the curve.✓ Proved
\[ - \frac{\frac{\partial}{\partial x} \left(x^{2} + y^{2}\right)}{\frac{\partial}{\partial y} \left(x^{2} + y^{2}\right)} = - \frac{x}{y} \]
y′ = −F_x/F_y.✓ Proved
\[ \frac{d}{d x} \left(- \frac{x}{Y{\left(x \right)}}\right) = \frac{x \frac{d}{d x} Y{\left(x \right)}}{Y^{2}{\left(x \right)}} - \frac{1}{Y{\left(x \right)}} \]
Differentiate y′ again, with y a function of x.✓ Proved
Replace dy/dx in that by y′ itself, and simplify.
Reviewed
\[ \left. \frac{- x^{2} - y^{2}}{y^{3}} \right|_{\substack{ x=-3\\ y=4 }} = - \frac{25}{64} \]
At (-3, 4).✓ Proved
Answer \( y'' = \frac{- x^{2} - y^{2}}{y^{3}},\quad y''(-3) = - \frac{25}{64} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The curve \( \displaystyle 3 x^{2} + y^{2} = 4 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle y'' \) there.
Show the first step
Show the whole solution
\[ \left. 3 x^{2} + y^{2} \right|_{\substack{ x=1\\ y=1 }} = 4 \]
The point is on the curve.✓ Proved
\[ - \frac{\frac{\partial}{\partial x} \left(3 x^{2} + y^{2}\right)}{\frac{\partial}{\partial y} \left(3 x^{2} + y^{2}\right)} = - \frac{3 x}{y} \]
y′ = −F_x/F_y.✓ Proved
\[ \frac{d}{d x} \left(- \frac{3 x}{Y{\left(x \right)}}\right) = \frac{3 x \frac{d}{d x} Y{\left(x \right)}}{Y^{2}{\left(x \right)}} - \frac{3}{Y{\left(x \right)}} \]
Differentiate y′ again, with y a function of x.✓ Proved
Replace dy/dx in that by y′ itself, and simplify.
Reviewed
\[ \left. - \frac{9 x^{2}}{y^{3}} - \frac{3}{y} \right|_{\substack{ x=1\\ y=1 }} = -12 \]
At (1, 1).✓ Proved
Answer \( y'' = - \frac{9 x^{2}}{y^{3}} - \frac{3}{y},\quad y''(1) = -12 \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The curve \( \displaystyle x^{2} + 4 y^{2} = 5 \) passes through \( \displaystyle (1, 1) \). Find \( \displaystyle y'' \) there.
Show the first step
Show the whole solution
\[ \left. x^{2} + 4 y^{2} \right|_{\substack{ x=1\\ y=1 }} = 5 \]
The point is on the curve.✓ Proved
\[ - \frac{\frac{\partial}{\partial x} \left(x^{2} + 4 y^{2}\right)}{\frac{\partial}{\partial y} \left(x^{2} + 4 y^{2}\right)} = - \frac{x}{4 y} \]
y′ = −F_x/F_y.✓ Proved
\[ \frac{d}{d x} \left(- \frac{x}{4 Y{\left(x \right)}}\right) = \frac{x \frac{d}{d x} Y{\left(x \right)}}{4 Y^{2}{\left(x \right)}} - \frac{1}{4 Y{\left(x \right)}} \]
Differentiate y′ again, with y a function of x.✓ Proved
Replace dy/dx in that by y′ itself, and simplify.
Reviewed
\[ \left. \frac{- x^{2} - 4 y^{2}}{16 y^{3}} \right|_{\substack{ x=1\\ y=1 }} = - \frac{5}{16} \]
At (1, 1).✓ Proved
Answer \( y'' = \frac{- x^{2} - 4 y^{2}}{16 y^{3}},\quad y''(1) = - \frac{5}{16} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The curve \( \displaystyle 2 x^{2} + 2 y^{2} = 10 \) passes through \( \displaystyle (-1, 2) \). Find \( \displaystyle y'' \) there.
Show the first step
Show the whole solution
\[ \left. 2 x^{2} + 2 y^{2} \right|_{\substack{ x=-1\\ y=2 }} = 10 \]
The point is on the curve.✓ Proved
\[ - \frac{\frac{\partial}{\partial x} \left(2 x^{2} + 2 y^{2}\right)}{\frac{\partial}{\partial y} \left(2 x^{2} + 2 y^{2}\right)} = - \frac{x}{y} \]
y′ = −F_x/F_y.✓ Proved
\[ \frac{d}{d x} \left(- \frac{x}{Y{\left(x \right)}}\right) = \frac{x \frac{d}{d x} Y{\left(x \right)}}{Y^{2}{\left(x \right)}} - \frac{1}{Y{\left(x \right)}} \]
Differentiate y′ again, with y a function of x.✓ Proved
Replace dy/dx in that by y′ itself, and simplify.
Reviewed
\[ \left. \frac{- x^{2} - y^{2}}{y^{3}} \right|_{\substack{ x=-1\\ y=2 }} = - \frac{5}{8} \]
At (-1, 2).✓ Proved
Answer \( y'' = \frac{- x^{2} - y^{2}}{y^{3}},\quad y''(-1) = - \frac{5}{8} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The curve \( \displaystyle x^{2} + 3 y^{2} = 4 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle y'' \) there.
Show the first step
Show the whole solution
\[ \left. x^{2} + 3 y^{2} \right|_{\substack{ x=1\\ y=-1 }} = 4 \]
The point is on the curve.✓ Proved
\[ - \frac{\frac{\partial}{\partial x} \left(x^{2} + 3 y^{2}\right)}{\frac{\partial}{\partial y} \left(x^{2} + 3 y^{2}\right)} = - \frac{x}{3 y} \]
y′ = −F_x/F_y.✓ Proved
\[ \frac{d}{d x} \left(- \frac{x}{3 Y{\left(x \right)}}\right) = \frac{x \frac{d}{d x} Y{\left(x \right)}}{3 Y^{2}{\left(x \right)}} - \frac{1}{3 Y{\left(x \right)}} \]
Differentiate y′ again, with y a function of x.✓ Proved
Replace dy/dx in that by y′ itself, and simplify.
Reviewed
\[ \left. \frac{- x^{2} - 3 y^{2}}{9 y^{3}} \right|_{\substack{ x=1\\ y=-1 }} = \frac{4}{9} \]
At (1, -1).✓ Proved
Answer \( y'' = \frac{- x^{2} - 3 y^{2}}{9 y^{3}},\quad y''(1) = \frac{4}{9} \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The curve \( \displaystyle x^{2} + y^{2} = 2 \) passes through \( \displaystyle (1, -1) \). Find \( \displaystyle y'' \) there.
Show the first step
Show the whole solution
\[ \left. x^{2} + y^{2} \right|_{\substack{ x=1\\ y=-1 }} = 2 \]
The point is on the curve.✓ Proved
\[ - \frac{\frac{\partial}{\partial x} \left(x^{2} + y^{2}\right)}{\frac{\partial}{\partial y} \left(x^{2} + y^{2}\right)} = - \frac{x}{y} \]
y′ = −F_x/F_y.✓ Proved
\[ \frac{d}{d x} \left(- \frac{x}{Y{\left(x \right)}}\right) = \frac{x \frac{d}{d x} Y{\left(x \right)}}{Y^{2}{\left(x \right)}} - \frac{1}{Y{\left(x \right)}} \]
Differentiate y′ again, with y a function of x.✓ Proved
Replace dy/dx in that by y′ itself, and simplify.
Reviewed
\[ \left. \frac{- x^{2} - y^{2}}{y^{3}} \right|_{\substack{ x=1\\ y=-1 }} = 2 \]
At (1, -1).✓ Proved
Answer \( y'' = \frac{- x^{2} - y^{2}}{y^{3}},\quad y''(1) = 2 \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify
The curve \( \displaystyle 4 x^{2} + y^{2} = 8 \) passes through \( \displaystyle (-1, 2) \). Find \( \displaystyle y'' \) there.
Show the first step
Show the whole solution
\[ \left. 4 x^{2} + y^{2} \right|_{\substack{ x=-1\\ y=2 }} = 8 \]
The point is on the curve.✓ Proved
\[ - \frac{\frac{\partial}{\partial x} \left(4 x^{2} + y^{2}\right)}{\frac{\partial}{\partial y} \left(4 x^{2} + y^{2}\right)} = - \frac{4 x}{y} \]
y′ = −F_x/F_y.✓ Proved
\[ \frac{d}{d x} \left(- \frac{4 x}{Y{\left(x \right)}}\right) = \frac{4 x \frac{d}{d x} Y{\left(x \right)}}{Y^{2}{\left(x \right)}} - \frac{4}{Y{\left(x \right)}} \]
Differentiate y′ again, with y a function of x.✓ Proved
Replace dy/dx in that by y′ itself, and simplify.
Reviewed
\[ \left. - \frac{16 x^{2}}{y^{3}} - \frac{4}{y} \right|_{\substack{ x=-1\\ y=2 }} = -4 \]
At (-1, 2).✓ Proved
Answer \( y'' = - \frac{16 x^{2}}{y^{3}} - \frac{4}{y},\quad y''(-1) = -4 \)
✓ Nihil obstat Lines: 4 proved, 1 reviewed. The answer was also checked a second way, without looking at the solution. Reviewers found nothing wrong with the explanation. The full receipt ·
How we verify