Implicit Differentiation (College Board AP® Calculus AB): Exam Questions

Consider the curve given by the equation $3y^2-12=y\cos x$ for $y>0$.

Show that $\frac{dy}{dx}=\frac{y\sin x}{\cos x-6y}$.

Write an equation for the line tangent to the curve at the point $\left(\frac{3\pi }{2},2\right)$.

A celestial body changes its temperature, $T$ degrees Fahrenheit, over time, $t$ minutes, according to the relationship:

$T^2+tT=500$

Find the rate at which the temperature of the body changes, $\frac{dT}{dt}$.

Is the celestial body cooling faster when it is $10°\mathrm{F}$ or when it is $20°\mathrm{F}?$ Explain your reasoning.

Consider the curve given by the equation $\frac{1}{5}{x}^2{e}^y=5.$

Find an expression for $\frac{dy}{dx}$ in terms of $x$ and $y.$

Find the slope of the tangents to the curve at the two points where the curve intersects the $x$-axis.

Show that the derivative of $y=\arcsin \left(3x\right)$ is $\frac{dy}{dx}=\frac{3}{\sqrt{1-9x^2}}$.

Show the work that leads to your answer.

Consider the curve given by the equation $\ln y+x{y}^2=1.$

Find an expression for $\frac{dy}{dx}$ in terms of $x$ and $y.$

The curve defined by $\ln y+x{y}^2=1$ passes through the point $\left(1,1\right).$

Write an equation for the line normal to the curve at the point $\left(1,1\right),$ express your final answer in slope-intercept form.

Is the horizontal line $y=1$ tangent to the curve $x^2+3y+2y^2=48$? Give a reason for your answer.

The curve $x^2+3y+2y^2=48$ intersects the positive $x$-axis at the point $\left(\sqrt{48}, 0\right)$. Is the line tangent to the curve at this point vertical? Give a reason for your answer.

Consider the function $y=f\left(x\right)$ whose curve is given by the equation $2y^2-6=y\sin x$ for $y>0$.

Show that $\frac{\text{d}y}{\text{d}x}=\frac{y\cos x}{4y-\sin x}$.

Consider the curve given by the equation $8xy=3+y^4$.

Show that $\frac{dy}{dx}=\frac{2y}{y^3-2x}$.

A particle is moving along the curve. At the instant when the particle is at the point $\left(\frac{7}{2},3\right)$, its horizontal position is increasing at a rate of $\frac{dx}{dt}=10$ units per second.

What is the value of $\frac{dy}{dt}$, the rate of change of the particle's vertical position, at that instant?

Use implicit differentiation to show that the derivative of $y=x^x$ is $\frac{dy}{dx}=x^x\left(\ln x+\hspace{0.17em}1\right)$.

Show all steps of your working.

Show that if $y=\arccos \left(x^{2}y\right)$ then $\frac{dy}{dx}=\frac{2xy}{-x^2-\sqrt{1-{\left(x^{2}y\right)}^2}}.$

A treasure chest is being raised from the bottom of the ocean using a pulley system. The depth of the chest below the ocean surface, $H,$ in meters, and the length of rope unwound from the pulley, $R,$ in meters, are related by the equation:

$H^3+tH=R^2+30$

where $t$ is the time elapsed, in minutes, since the pulley started operating.

Show that $\frac{dH}{dt}=\frac{2R{\displaystyle \frac{dR}{dt}}-H}{3H^2+t}$ and explain what this represents in the context of the problem.

At $t=4$ minutes, the chest is at a depth of $3$meters and the length of the rope is changing at a rate of $-1$meters per minute.

Find the rate at which the depth of the treasure chest is changing at $t=4$ minutes. Explain whether this means the treasure chest is rising or falling at this time.

At $t=5$ minutes the length of the rope is now $1$ meter and the rate at which the length of rope is changing has not changed. Determine whether the depth of the chest is changing faster or slower than at $t=4$ minutes.

Consider the curve given by the equation $\ln \left(xy\right)+x{y}^2=1.$

Show that $\frac{dy}{dx}=\frac{-y-x{y}^3}{x+2x^2{y}^2}$.

Verify that the point with coordinates $\left(1,1\right)$ lies on the curve with equation $\ln \left(xy\right)+x{y}^2=1.$

Then, find the equation of the tangent to the curve at this point. Express your final answer in slope-intercept form.

The tangent line to the curve with equation $\ln \left(xy\right)+x{y}^2=1$ at the point with coordinates $\left(1,1\right)$ intercepts the $x$-axis at the point $A$ and the $y$-axis at the point $B.$ Find the area of the triangle formed by the origin, $O$ and the points $A$ and $B.$

Consider the curve given by the equation $6xy=2+y^3$.

Show that $\frac{\text{d}y}{\text{d}x}=\frac{2y}{y^2-2x}$.

Find the coordinates of a point on the curve at which the line tangent to the curve is horizontal, or explain why no such point exists.

Find the coordinates of a point on the curve at which the line tangent to the curve is vertical, or explain why no such point exists.

Consider the curve $G$ defined by the equation $y^3-y^2-y+\frac{1}{4}{x}^2=0$.

Show that $\frac{\text{d}y}{\text{d}x}=\frac{-x}{2\left(3y^2-2y-1\right)}$.