Show that $\frac{dy}{dx}=\pm \frac{3x^2+1}{2\sqrt{x^3+x}}$, for $x>0$.

Find the rate of change of the height of the water when the container is filled to half its maximum volume.

Find the coordinates of the three points on C, nearest the origin, where the tangent is parallel to the line $y=-x$.

Using implicit differentiation, find an expression for $\frac{\text{d}y}{\text{d}x}$.

Find the equation of the tangent to the curve at the point $\left(\frac{1}{4}\text{, }\frac{5\pi }{6}\right)$.

Show that $\frac{\text{d}y}{\text{d}x}=-\left(\frac{1+y\text{sin}\left(xy\right)}{1+x\text{sin}\left(xy\right)}\right)$.

Find $\frac{\text{d}y}{\text{d}x}$ in terms of $x$ and $y$.

Find the time it takes to fill the container to its maximum volume.

Find the coordinates of the points on the curve $y^3+3x{y}^2-x^3=27$ at which $\frac{\text{d}y}{\text{d}x}=0$.

The curve has a point of inflexion at $\left(x_1, y_1\right)$ where ${\text{e}}^{-\frac{\mathrm{\pi }}{2}}<x_1<{\text{e}}^{\frac{\mathrm{\pi }}{2}}$. Determine the coordinates of this point of inflexion.

Find the equations of the tangents to this curve at the points where the curve intersects the line $x=1$.

Calculate $\frac{\text{d}\theta }{\text{d}t}$ when $\theta =\frac{\pi }{3}$.

The folium of Descartes is a curve defined by the equation $x^3+y^3-3xy=0$, shown in the following diagram.

Determine the exact coordinates of the point P on the curve where the tangent line is parallel to the $y$-axis.

Show that $\frac{\text{d}y}{\text{d}x}=\frac{y \cos \left(xy\right)}{2y-x \cos \left(xy\right)}$.

Let P($a$, $b$) be the unique point where the curves $C_1$ and $C_2$ intersect.

Show that the tangent to $C_1$ at P is perpendicular to the tangent to $C_2$ at P.

Prove that, when $\frac{\text{d}y}{\text{d}x}=0 , y=\pm 1$.

The tangent to $C$ at the point Ρ is parallel to the $y$-axis.

Find the $x$-coordinate of Ρ.

Hence find the equation of the tangent to $C$ at the point where $x=1$.

Find the equation of the tangent to $C$ at $\text{P}$.

By substituting $Y=\frac{dy}{dt}$, show that $Y=B{\text{e}}^{2t}$ where $B$ is a constant.

Let the two values found in part (c)(ii) be ${\lambda }_1$ and ${\lambda }_2$.

Verify that $y=F{\text{e}}^{{\lambda }_{1}t}+G{\text{e}}^{{\lambda }_{2}t}$ is a solution to the differential equation in (c)(i),where $G$ is a constant.

Find an expression for the volume of water $V({\text{m}}^3)$ in the trough in terms of $\theta$.

Determine the equation of the tangent to $C$ at the point $\left(\frac{2}{\text{e}},\text{ e}\right)$

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

Hence find the coordinates of all points on $C$, for $0<x<4\mathrm{\pi }$, where $\frac{\text{d}y}{\text{d}x}=0$.

Find the coordinates of P and Q.

Using implicit differentiation, or otherwise, find $\frac{\text{d}y}{\text{d}x}$ for each curve in terms of $x$ and $y$.

Given that the gradients of the tangents to C at P and Q are m₁ and m₂ respectively, show that m₁ × m₂ = 1.

Show that $\frac{dy}{dx}=\frac{3x^2}{2{\text{e}}^{2y}-1}$.

Use the differential equation $\frac{dy}{dx}=\frac{y^2+3xy+2x^2}{x^2}$ to show that the points of zero gradient on the curve lie on two straight lines of the form $y=mx$ where the values of $m$ are to be determined.

At time $T$, the following conditions are true.

Boat $\text{B}$ has travelled $10$ metres further than boat $\text{A}$.
Boat $\text{B}$ is travelling at double the speed of boat $\text{A}$.
The rate of change of the angle $\theta$ is $-0.1$ radians per second.

Find the speed of boat $\text{A}$ at time $T$.

Show that $\frac{dy}{dx}+\left(x\frac{{\displaystyle dy}}{{\displaystyle dx}}+y\right)\left(1+\ln \left(xy\right)\right)=1$.

By solving the differential equation $\frac{dy}{dt}=y$, show that $y=A{\text{e}}^t$ where $A$ is a constant.

Show that $\frac{dx}{dt}-x=-A{\text{e}}^t$.

Solve the differential equation in part (a)(ii) to find $x$ as a function of $t$.

By differentiating $\frac{dy}{dt}=-x+y$ with respect to $t$, show that $\frac{d^{2}y}{d{t}^2}=2\frac{{\displaystyle dy}}{{\displaystyle dt}}$.

Hence find $y$ as a function of $t$.

Hence show that $x=-\frac{B}{2}{\text{e}}^{2t}+C$, where $C$ is a constant.

Show that $\frac{d^{2}y}{d{t}^2}-2\frac{dy}{dt}-3y=0$.

Find the two values for $\lambda$ that satisfy $\frac{d^{2}y}{d{t}^2}-2\frac{dy}{dt}-3y=0$.