Free math worksheets

- Differentiate both variables with respect to \(x\), remembering that \(y\) depends on \(x\). - A horizontal tangent and a vertical tangent correspond to different parts of the slope fraction. - Substitute each resulting condition back into the original relation.

1. Implicit differentiation gives \(2x+2y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{x}{y}\). 2. A horizontal tangent requires \(x=0\) with \(y\ne0\). The relation gives \((0, 5)\) and \((0, -5)\). 3. A vertical tangent occurs where the denominator is zero while the numerator is nonzero: \(y=0\), giving \((5, 0)\) and \((-5, 0)\).

Horizontal tangents: \((0, 5)\) and \((0, -5)\). Vertical tangents: \((5, 0)\) and \((-5, 0)\).

- Differentiate every occurrence of \(y\) as a function of \(x\). - The product \(xy\) needs two derivative terms. - Substitute the point only after isolating the slope.

1. Differentiate implicitly: \(2x+y+x\frac{dy}{dx}=0\). 2. Solve for the derivative: \(\frac{dy}{dx}=-\frac{2x+y}{x}\). 3. At \((2, 1)\), the slope is \(-\frac{5}{2}\). 4. The tangent line is \(y-1=-\frac52(x-2)\).

Slope: \(-\frac52\). Tangent line: \(y-1=-\frac52(x-2)\).

- First obtain the slope as a fraction from implicit differentiation. - Horizontal and vertical tangents come from different zero conditions. - The top and bottom of the relation occur at the horizontal-tangent points.

1. Differentiate implicitly: \(2(x-1)+2y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{x-1}{y}\). 2. Horizontal tangents require \(x=1\). The relation gives \((1, 2)\) and \((1, -2)\). 3. Vertical tangents require \(y=0\). The relation gives \((-1, 0)\) and \((3, 0)\). 4. The highest y-value is \(2\), and the lowest y-value is \(-2\).

Horizontal tangents: \((1, 2)\), \((1, -2)\). Vertical tangents: \((-1, 0)\), \((3, 0)\). Highest y-value: \(2\); lowest y-value: \(-2\).

- Interpret a derivative written as a fraction by checking numerator and denominator separately. - A zero numerator and a zero denominator do not mean the same thing. - When both are zero, avoid canceling or assigning a slope without additional information.

1. At \(A\), the numerator is zero and the denominator is nonzero, so the tangent is horizontal. 2. At \(B\), the denominator is zero and the numerator is nonzero, so the slope is not finite; this indicates a vertical tangent in the ordinary smooth case. 3. At \(C\), the slope is \(\frac{2}{-4}=-\frac12\). 4. At \(D\), both numerator and denominator are zero, so the derivative formula alone is inconclusive. More analysis of the relation is required.

\(A\): horizontal tangent. \(B\): vertical tangent in the ordinary smooth case. \(C\): slope \(-\frac12\). \(D\): inconclusive from the derivative fraction alone.

- Isolate the slope after differentiating the relation implicitly. - Horizontal and vertical tangents impose different conditions on the resulting fraction. - Combine each slope condition with the original relation to recover complete points.

1. Implicit differentiation gives \(2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{2x+y}{x+2y}\). 2. A horizontal tangent requires \(2x+y=0\). Substituting \(y=-2x\) gives \(3x^2=7\), so the horizontal-tangent points are \(\left(\frac{\sqrt{21}}{3}, -\frac{2\sqrt{21}}{3}\right)\) and \(\left(-\frac{\sqrt{21}}{3}, \frac{2\sqrt{21}}{3}\right)\). 3. A vertical tangent requires \(x+2y=0\). Substituting \(x=-2y\) gives \(3y^2=7\), yielding \(\left(-\frac{2\sqrt{21}}{3}, \frac{\sqrt{21}}{3}\right)\) and \(\left(\frac{2\sqrt{21}}{3}, -\frac{\sqrt{21}}{3}\right)\).

Horizontal tangents: \(\left(\frac{\sqrt{21}}{3}, -\frac{2\sqrt{21}}{3}\right)\), \(\left(-\frac{\sqrt{21}}{3}, \frac{2\sqrt{21}}{3}\right)\). Vertical tangents: \(\left(-\frac{2\sqrt{21}}{3}, \frac{\sqrt{21}}{3}\right)\), \(\left(\frac{2\sqrt{21}}{3}, -\frac{\sqrt{21}}{3}\right)\).

- Differentiate the original relation once before substituting the point. - For the second derivative, differentiate the entire first differentiated relation again. - The sign of the second derivative determines local concavity.

1. Differentiate implicitly: \(2x+8y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{x}{4y}\). At \((2, \sqrt3)\), \(\frac{dy}{dx}=-\frac{1}{2\sqrt3}\). 2. Differentiate \(2x+8y\frac{dy}{dx}=0\) again: \(2+8\left[\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}\right]=0\). 3. Substitute \(y=\sqrt3\) and \(\left(\frac{dy}{dx}\right)^2=\frac1{12}\). This gives \(\frac{d^2y}{dx^2}=-\frac{1}{3\sqrt3}\). 4. The second derivative is negative, so the branch is concave down at the point.

\(\frac{dy}{dx}=-\frac{1}{2\sqrt3}\), \(\frac{d^2y}{dx^2}=-\frac{1}{3\sqrt3}\); the branch is concave down.

- Use the horizontal-tangent information when simplifying the second derivative. - Keep \(y\) symbolic until you substitute each point. - Interpret the sign separately on the upper and lower branches.

1. Implicit differentiation gives \(2x+4y\frac{dy}{dx}=0\). 2. Differentiate again: \(2+4\left[\left(\frac{dy}{dx}\right)^2+y\frac{d^2y}{dx^2}\right]=0\). 3. At either given point, \(\frac{dy}{dx}=0\), so \(\frac{d^2y}{dx^2}=-\frac{1}{2y}\). 4. At \((0, 2)\), \(\frac{d^2y}{dx^2}=-\frac14<0\), so the upper branch has a local maximum. 5. At \((0, -2)\), \(\frac{d^2y}{dx^2}=\frac14>0\), so the lower branch has a local minimum.

\((0, 2)\) is a local maximum of the upper branch. \((0, -2)\) is a local minimum of the lower branch.

- Look for a term that is a product of two quantities that both vary with \(x\). - Collect all terms containing \(\frac{dy}{dx}\) before solving for the slope. - A horizontal tangent provides a condition to combine with the original relation.

1. The term \(2xy\) requires the product rule: \(\frac{d}{dx}(2xy)=2y+2x\frac{dy}{dx}\). Thus \(2x+2y+2x\frac{dy}{dx}+6y\frac{dy}{dx}=0\). 2. Solving gives \(\frac{dy}{dx}=-\frac{x+y}{x+3y}\). 3. A horizontal tangent requires \(x+y=0\). With \(y=-x\), the relation becomes \(2x^2=12\), so \(x=\pm\sqrt6\). 4. The points are \((\sqrt6, -\sqrt6)\) and \((-\sqrt6, \sqrt6)\).

a) The product-rule term \(2x\frac{dy}{dx}\) was omitted. The correct derivative is \(\frac{dy}{dx}=-\frac{x+y}{x+3y}\). b) \((\sqrt6, -\sqrt6)\) and \((-\sqrt6, \sqrt6)\).

- Use the sign of the implicit derivative rather than solving explicitly for \(y\). - The branch condition tells you the sign of the denominator. - Respect the x-values for which the relation actually has points on the chosen branch.

1. Implicit differentiation gives \(2x-2y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=\frac{x}{y}\). 2. On the upper branch, \(y>0\), so the sign of \(\frac{dy}{dx}\) is the sign of \(x\). 3. The upper branch has points only for \(x<-1\) or \(x>1\). Therefore, it is decreasing on \((-\infty, -1)\) and increasing on \((1, \infty)\).

\(\frac{dy}{dx}=\frac{x}{y}\). Upper branch decreasing on \((-\infty, -1)\); increasing on \((1, \infty)\).

- Differentiate both sides while treating \(y\) as a function of \(x\). - Collect all terms containing \(\frac{dy}{dx}\) before substituting the point. - Simplify the numerator and denominator separately at the given point.

1. Differentiate implicitly: \(3x^2+3y^2\frac{dy}{dx}=6y+6x\frac{dy}{dx}\). 2. Collect derivative terms: \((3y^2-6x)\frac{dy}{dx}=6y-3x^2\). 3. At \((3, 3)\), \(\frac{dy}{dx}=\frac{18-27}{27-18}=-1\).

The tangent slope at \((3, 3)\) is \(-1\).

- Local extrema of a branch occur where its tangent is horizontal. - Use the original relation to recover the y-coordinates after finding the x-condition. - The second derivative distinguishes the upper and lower branch behavior.

1. Differentiate implicitly: \(2x+6y\frac{dy}{dx}-4=0\), so \(\frac{dy}{dx}=\frac{2-x}{3y}\). 2. Horizontal tangents require \(x=2\). Substituting into the relation gives \(4+3y^2-8=0\), so \(y=\pm\frac{2}{\sqrt3}\). 3. Differentiate \(2x+6yy'-4=0\) again: \(2+6[(y')^2+yy'']=0\). 4. At a horizontal tangent, \(y'=0\), so \(y''=-\frac{1}{3y}\). 5. At \(\left(2, \frac{2}{\sqrt3}\right)\), \(y''<0\), giving a local maximum. At \(\left(2, -\frac{2}{\sqrt3}\right)\), \(y''>0\), giving a local minimum.

Local maximum: \(\left(2, \frac{2}{\sqrt3}\right)\). Local minimum: \(\left(2, -\frac{2}{\sqrt3}\right)\).

- A derivative equals zero only when its numerator is zero and its denominator remains defined. - Use the original relation to see whether the numerator condition can force the denominator to zero as well. - Check the denominator at the actual points, not only symbolically in general.

1. The implicit derivative is \(\frac{dy}{dx}=-\frac{2x+y}{x+2y}\). 2. A horizontal tangent requires the numerator to be zero and the denominator to be nonzero. 3. From \(2x+y=0\), \(y=-2x\), and the relation gives \(3x^2=7\), so \(x\ne0\). 4. At these points, the denominator is \(x+2(-2x)=-3x\ne0\). Therefore, both points found from the numerator condition genuinely have horizontal tangents.

The denominator must also be nonzero. At the points satisfying \(y=-2x\) on the relation, \(x=\pm\sqrt{7/3}\ne0\), so \(x+2y=-3x\ne0\); both are valid horizontal-tangent points.

- Keep the differentiated relation unsolved long enough to inspect what happens at the point. - A vertical tangent corresponds to an undefined finite slope, not to a zero slope. - Whenever an algebraic step divides by an expression, check separately where that expression is zero.

1. Differentiate implicitly: \(2y\frac{dy}{dx}=3x^2-1\). 2. At \((1, 0)\), the left coefficient of \(\frac{dy}{dx}\) is \(0\), while the right side is \(2\ne0\). Thus no finite derivative can satisfy the equation, which corresponds to a vertical tangent. 3. Solving formally gives \(\frac{dy}{dx}=\frac{3x^2-1}{2y}\), whose denominator is zero at \((1, 0)\) while the numerator is nonzero. 4. Dividing by \(y\) before checking \(y=0\) can discard points where the slope is undefined and therefore miss vertical tangents.

\((1, 0)\) has a vertical tangent because \(2y=0\) while \(3x^2-1=2\ne0\). Dividing by \(y\) without separately checking \(y=0\) can discard such points.

- Obtain the first derivative from the implicit relation before differentiating a second time. - In the second differentiation, remember that both \(y\) and \(y'\) depend on \(x\). - Substitute the known point and first-derivative value only after forming the second-derivative equation.

1. Differentiate: \(2x+y+x\frac{dy}{dx}+2y\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{2x+y}{x+2y}\). At \((1, 1)\), \(\frac{dy}{dx}=-1\). 2. Differentiate \(2x+y+(x+2y)y'=0\): \(2+y'+(1+2y')y'+(x+2y)y''=0\). 3. At \((1, 1)\) with \(y'=-1\), the equation becomes \(2-1+(1-2)(-1)+3y''=0\), so \(2+3y''=0\). 4. Thus \(y''=-\frac23<0\), so the branch is concave down at \((1, 1)\).

\(\frac{dy}{dx}=-1\), \(\frac{d^2y}{dx^2}=-\frac23\); the branch is concave down at \((1, 1)\).

- Factor the quadratic expression before treating the relation as a generic conic. - Compare the factored form with the differentiated equation. - Once the components are recognized, interpret what their constant slopes imply.

1. The left side is \((x+2y)^2\), so the relation is \(x+2y=3\) or \(x+2y=-3\). It is the union of two parallel lines. 2. Implicit differentiation gives \(2x+4y+4x y'+8y y'=0\). Factoring gives \(2(x+2y)(1+2y')=0\). 3. On either line, \(x+2y=\pm3\ne0\), so \(1+2y'=0\), giving \(y'=-\frac12\). 4. Every point on either component has the same tangent slope \(-\frac12\); there are no horizontal or vertical tangents.

a) The relation is the pair of lines \(x+2y=\pm3\). b) The differentiated equation factors because the original relation is degenerate rather than an ellipse. c) The tangent slope is \(-\frac12\) everywhere on both lines.