- Differentiate both sides with respect to \(x\). - Treat \(y\) as a function of \(x\), not as a constant. - The derivative of the constant on the right side is \(0\).

1. Differentiate both sides with respect to \(x\): \(1+\frac{dy}{dx}=0\). 2. Therefore, \(\frac{dy}{dx}=-1\).

\(\frac{dy}{dx}=-1\)

- Substitute the point into the differentiated equation first. - After substitution, \(\frac{dy}{dx}\) is the only unknown. - Solve the resulting linear equation for the slope.

1. Substitute \(x=3\) and \(y=4\): \(6+8\frac{dy}{dx}=0\). 2. Solving gives \(\frac{dy}{dx}=-\frac{3}{4}\).

\(-\frac{3}{4}\)

- Imagine temporarily writing \(y=y(x)\). - Differentiating the outer cube is only the first part of the chain rule.

1. Treat \(y\) as a function of \(x\), so \(y^3\) is a composite function. 2. The chain rule gives \(\frac{d}{dx}(y^3)=3y^2\frac{dy}{dx}\).

\(3y^2\frac{dy}{dx}\)

- First recall the derivative of cosine with an ordinary input. - What extra factor appears because the input is \(y(x)\)?

1. The derivative of cosine with respect to its input is negative sine. 2. Since its input is \(y(x)\), multiply by \(\frac{dy}{dx}\) to get \(-\sin y\frac{dy}{dx}\).

\(-\sin y\frac{dy}{dx}\)

- Which expression is being squared? - That expression is not constant as x changes.

1. Differentiating the outside square gives \(2(y+1)\). 2. The inner expression \(y+1\) changes at rate \(\frac{dy}{dx}\), so the derivative is \(2(y+1)\frac{dy}{dx}\).

The missing factor is \(\frac{dy}{dx}\).

- What operation connects \(x\) and \(y^2\)? - What changes when you differentiate a power of \(y(x)\) instead of a power of \(x\)?

1. The left side multiplies \(x\) by \(y^2\), so the product rule applies to the whole expression. 2. Differentiating the factor \(y^2\) also needs the chain rule because \(y\) depends on \(x\).

Use the product rule on \(x\cdot y^2\) and the chain rule when differentiating \(y^2\).

- Treat \(y\) as a function whose input is \(x\), not as a constant. - For a power of \(y\), think of the power as the outer function and \(y(x)\) as the inner function. - When \(x\) and \(y\) are multiplied, both factors depend on the differentiation process.

1. a) Because \(y=y(x)\), use the chain rule: \(\frac{d}{dx}(y^2)=2y y^{\prime}\). The factor \(y^{\prime}\) comes from differentiating the inner function \(y(x)\). 2. b) Differentiate term by term: \(\frac{d}{dx}(3x+y^3)=3+3y^2y^{\prime}\). The factor \(y^{\prime}\) comes from the chain rule on \(y^3\). 3. c) Use the product rule: \(\frac{d}{dx}(xy)=y+xy^{\prime}\). The factor \(y^{\prime}\) appears when differentiating the factor \(y(x)\).

a) \(2y y^{\prime}\); chain rule b) \(3+3y^2y^{\prime}\); chain rule on \(y^3\) c) \(y+xy^{\prime}\); product rule, with \(y^{\prime}\) from differentiating \(y(x)\)

- Differentiate both sides of the equation with respect to \(x\). - When differentiating a power of \(y\), remember that \(y\) depends on \(x\). - Isolate \(\frac{dy}{dx}\) before substituting the coordinates of the point.

1. a) Differentiate both sides with respect to \(x\): \(2x+4y\frac{dy}{dx}=0\). Solve for the derivative: \(\frac{dy}{dx}=-\frac{x}{2y}\). 2. b) The point lies on the curve because \(4^2+2(1)^2=18\). Substitute \(x=4\) and \(y=1\): \(\frac{dy}{dx}=-\frac{4}{2(1)}=-2\).

a) \(\frac{dy}{dx}=-\frac{x}{2y}\) b) \(-2\)

- Treat \(y\) as changing with \(x\) when you differentiate \(y^2\). - Find a formula for the slope before substituting the point.

1. Since \(y\) depends on \(x\), differentiating both sides gives \(2x+2y\frac{dy}{dx}=0\). 2. Rearrange to \(\frac{dy}{dx}=-\frac{x}{y}\) for \(y\ne0\). 3. At \((1,2)\), the slope is \(-\frac{1}{2}\).

\(\frac{dy}{dx}=-\frac{x}{y}\); slope at \((1,2)\): \(-\frac{1}{2}\).

- Differentiate both terms with respect to x, remembering that y depends on x. - Collect the term containing \(\frac{dy}{dx}\) before dividing. - What happens to that division when y is zero?

1. Differentiate both sides: \(2x+2y\frac{dy}{dx}=0\). 2. When \(y\ne 0\), isolate the derivative: \(\frac{dy}{dx}=-\frac{x}{y}\). 3. At \((\sqrt{10},0)\), the coefficient of \(\frac{dy}{dx}\) in the differentiated equation is zero. A finite slope cannot satisfy \(2\sqrt{10}+0=0\); the circle has a vertical tangent there.

\(\frac{dy}{dx}=-\frac{x}{y}\) when \(y\ne 0\). At \((\sqrt{10},0)\), the tangent is vertical, so the derivative does not exist.

- The exponent is y, which changes with x. - What extra factor appears when differentiating an exponential with an input y(x)? - Substitute the point only after isolating the derivative.

1. Differentiate both sides: \(e^y\frac{dy}{dx}+2=0\). 2. Isolate \(\frac{dy}{dx}=-\frac{2}{e^y}\). 3. At \((2,0)\), \(e^y=1\), so the slope is \(-2\).

\(\frac{dy}{dx}=-2\) at \((2,0)\).

- What information do you need from the plotted points before finding their slopes? - When differentiating \(xy\), remember that \(y\) depends on \(x\). - Compare the absolute values of the two slopes after substituting the coordinates.

1. Read \(P=(1,4)\) and \(Q=(4,1)\) from the graph. Both satisfy \(xy=4\). 2. Implicit differentiation gives \(y+x\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{y}{x}\). 3. The slopes are \(-4\) at \(P\) and \(-\frac{1}{4}\) at \(Q\). The tangent at \(P\) is steeper because its slope has greater absolute value.

a) \(P=(1,4)\), \(Q=(4,1)\) b) \(\frac{dy}{dx}=-\frac{y}{x}\); slopes \(-4\) and \(-\frac{1}{4}\), respectively. The tangent at \(P\) is steeper.

- Treat the term \(xy\) as a product of two factors before differentiating. - Keep every term containing \(\frac{dy}{dx}\) together after differentiating. - Substitute the point only after you have a derivative formula in terms of \(x\) and \(y\).

1. a) Differentiate both sides with respect to \(x\). The product rule gives \(\frac{d}{dx}(xy)=y+x\frac{dy}{dx}\), and the chain rule gives \(\frac{d}{dx}(y^2)=2y\frac{dy}{dx}\). Thus, \(y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\). Collect the derivative terms and solve: \(\frac{dy}{dx}=-\frac{y}{x+2y}\). 2. b) At \((1,2)\), \(\frac{dy}{dx}=-\frac{2}{1+4}=-\frac{2}{5}\). 3. c) Both \(x\) and \(y=y(x)\) vary with \(x\), so the product rule differentiates each factor in turn. The missing term would be \(x\frac{dy}{dx}\).

a) \(\frac{dy}{dx}=-\frac{y}{x+2y}\) b) \(-\frac{2}{5}\) c) Both factors in \(xy\) vary with \(x\), so the product rule gives \(y+x\frac{dy}{dx}\), not just \(y\).

- Differentiate the equation before substituting the coordinates of the point. - A power of \(y\) requires a chain-rule factor involving \(\frac{dy}{dx}\). - Once you have the slope, use the given point to write a tangent-line equation.

1. Differentiate implicitly: \(2x+3y^2\frac{dy}{dx}=0\). 2. Solve for the derivative: \(\frac{dy}{dx}=-\frac{2x}{3y^2}\). 3. At \((1,2)\), the slope is \(-\frac{2}{3(4)}=-\frac{1}{6}\). 4. Use point-slope form: \(y-2=-\frac{1}{6}(x-1)\).

\(y-2=-\frac{1}{6}(x-1)\)

- Ask whether \(y\) is being treated as a constant or as a function of \(x\). - When a trigonometric function has \(y\) as its input, differentiate the outer trig function and then the inner function. - Check that the given point satisfies the original equation before evaluating the derivative.

1. Matej treated \(y\) as if it were independent of \(x\). Since \(y=y(x)\), differentiating \(\sin y\) requires the chain rule and a factor of \(\frac{dy}{dx}\). 2. The correct differentiated equation is \(2x+\cos y\frac{dy}{dx}=0\). Therefore, \(\frac{dy}{dx}=-\frac{2x}{\cos y}\), wherever \(\cos y\ne0\). 3. The point lies on the curve because \(1^2+\sin0=1\). At \((1,0)\), \(\frac{dy}{dx}=-\frac{2}{1}=-2\).

a) The derivative of \(\sin y\) is missing the chain-rule factor \(\frac{dy}{dx}\). b) \(\frac{dy}{dx}=-\frac{2x}{\cos y}\) c) \(-2\)

- Use the grid and marker to determine the point before differentiating. - Differentiate both squared terms with respect to \(x\), remembering that \(y=y(x)\). - Use the derivative value as the tangent slope in point-slope form.

1. a) From the grid, \(P=(2,2)\). 2. b) Differentiate the curve equation: \(2x+8y\frac{dy}{dx}=0\). Thus \(\frac{dy}{dx}=-\frac{x}{4y}\). At \((2,2)\), the slope is \(-\frac{2}{8}=-\frac{1}{4}\). 3. c) Using point-slope form, the tangent line is \(y-2=-\frac{1}{4}(x-2)\).

a) \((2,2)\) b) \(-\frac{1}{4}\) c) \(y-2=-\frac{1}{4}(x-2)\)

- First find the slope formula in x and y. - A slope of 1 gives a relationship between x and y. - Substitute that relationship into the ellipse equation and check both solutions.

1. Differentiate: \(2x+8y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{x}{4y}\) for \(y\ne 0\). 2. Set the slope equal to \(1\): \(x=-4y\). 3. Substitute in the ellipse: \(16y^2+4y^2=20\), so \(y=1\) or \(y=-1\). 4. Then \(x=-4\) or \(x=4\), respectively. Both points lie on the ellipse.

\((-4,1)\) and \((4,-1)\)

- First solve the differentiated equation for \(\frac{dy}{dx}\). - A horizontal tangent is controlled by the numerator of the derivative, provided the denominator is nonzero. - A vertical tangent is controlled by the denominator, provided the numerator is nonzero. - Check the original curve equation after imposing each tangent condition.

1. Differentiate implicitly: \(2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\). Thus \(\frac{dy}{dx}=-\frac{2x+y}{x+2y}\). 2. a) A horizontal tangent requires \(2x+y=0\) and \(x+2y\ne0\). Set \(y=-2x\). Substitution into the curve gives \(3x^2=7\), so \(x=\pm\sqrt{\frac{7}{3}}\). The points are \(\left(\sqrt{\frac{7}{3}},-2\sqrt{\frac{7}{3}}\right)\) and \(\left(-\sqrt{\frac{7}{3}},2\sqrt{\frac{7}{3}}\right)\). At these points, \(x+2y=-3x\ne0\). 3. b) A vertical tangent requires \(x+2y=0\) and \(2x+y\ne0\). Set \(x=-2y\). Substitution gives \(3y^2=7\), so \(y=\pm\sqrt{\frac{7}{3}}\). The points are \(\left(-2\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\right)\) and \(\left(2\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}\right)\). At these points, \(2x+y=-3y\ne0\).

a) \(\left(\sqrt{\frac{7}{3}},-2\sqrt{\frac{7}{3}}\right)\) and \(\left(-\sqrt{\frac{7}{3}},2\sqrt{\frac{7}{3}}\right)\) b) \(\left(-2\sqrt{\frac{7}{3}},\sqrt{\frac{7}{3}}\right)\) and \(\left(2\sqrt{\frac{7}{3}},-\sqrt{\frac{7}{3}}\right)\)

- The parameter is constant when differentiating with respect to \(x\). - Find the implicit derivative before substituting the fixed point. - Parallel lines have equal slopes, so translate the line condition into an equation involving the parameter.

1. Differentiate implicitly with respect to \(x\): \(2x+2y\frac{dy}{dx}+a=0\). 2. Solve for the slope: \(\frac{dy}{dx}=-\frac{2x+a}{2y}\). 3. At \((1,2)\), the slope is \(-\frac{2+a}{4}\). 4. A line parallel to \(y=-x+6\) has slope \(-1\). Set \(-\frac{2+a}{4}=-1\). Then \(2+a=4\), so \(a=2\).

\(a=2\)

- Differentiate every occurrence of \(y\) as a function of \(x\). - After differentiating, collect all terms containing \(\frac{dy}{dx}\) on one side. - For the method comparison, focus on what information is needed to find a slope at one point rather than a formula for the entire curve.

1. a) Differentiate both sides with respect to \(x\): \(3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}\). Collect the derivative terms: \(\left(3y^2-6x\right)\frac{dy}{dx}=6y-3x^2\). Thus \(\frac{dy}{dx}=\frac{6y-3x^2}{3y^2-6x}\). At \((3,3)\), \(\frac{dy}{dx}=\frac{18-27}{27-18}=-1\). 2. b) Implicit differentiation works directly with the relation that the point satisfies and does not require choosing or solving for a global explicit branch. At \((3,3)\), the coefficient \(3y^2-6x=9\ne0\), so the differentiated equation determines a unique finite slope there.

a) \(-1\) b) Implicit differentiation uses the given relation directly and avoids solving a cubic for \(y\); at \((3,3)\), the derivative equation has a nonzero coefficient on \(\frac{dy}{dx}\), so it determines the local slope.