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| Logarithmic Differentiation (for product, quotient, and function power function forms) | Functions in Parametric Forms | Derivatives of Functions in Parametric Forms |
Differentiation Techniques: Logarithmic and Parametric
Logarithmic Differentiation (for product, quotient, and function power function forms)
Logarithmic differentiation is a useful technique for finding the derivative of functions that have a complicated structure involving products, quotients, and powers, especially when dealing with functions where both the base and the exponent are variables (function raised to the power of a function).
The method involves taking the natural logarithm of the function first. This simplifies the expression using the properties of logarithms. Then, we differentiate the resulting equation implicitly with respect to the independent variable (usually $x$), and finally solve for the derivative $\frac{dy}{dx}$.
This technique is particularly advantageous for:
- Functions that are products or quotients of many terms, or terms raised to powers. Using the Product and Quotient Rules repeatedly would be very cumbersome. Logarithms turn products into sums and quotients into differences, simplifying the differentiation process.
- Functions of the form $y = [f(x)]^{g(x)}$, where both the base $f(x)$ and the exponent $g(x)$ are non-constant functions of $x$. The standard Power Rule ($\frac{d}{dx}(x^n)=nx^{n-1}$, where $n$ is constant) and the standard Exponential Rule ($\frac{d}{dx}(a^x)=a^x \ln a$, where $a$ is constant) do not directly apply to this form.
The Process of Logarithmic Differentiation
To find the derivative of a function $y = f(x)$ using logarithmic differentiation, follow these steps:
- Take the Natural Logarithm of both sides. Start with the equation $y = f(x)$ and take the natural logarithm of both sides: $\ln y = \ln[f(x)]$. This step is valid if $y$ and $f(x)$ are positive. If $f(x)$ can take negative values, we might consider the derivative of $\ln|y| = \ln|f(x)|$. For finding $\frac{dy}{dx}$, assuming $y=f(x)$ represents a differentiable curve, we typically work on intervals where $f(x)$ is positive or negative and handle points where $f(x)=0$ separately if necessary.
- Simplify the right side using Logarithm Properties. Use the properties of natural logarithms to expand and simplify the expression $\ln[f(x)]$. The key properties are:
- $\ln(ab) = \ln a + \ln b$ (Product turns into Sum)
- $\ln(a/b) = \ln a - \ln b$ (Quotient turns into Difference)
- $\ln(a^n) = n \ln a$ (Power comes down as a factor)
- Differentiate both sides implicitly with respect to $x$. Differentiate each term on both sides of the simplified equation with respect to the independent variable $x$. Remember that $y$ is a function of $x$. When differentiating a term involving $y$, use the Chain Rule. The derivative of $\ln y$ with respect to $x$ is always $\frac{d}{dx}(\ln y) = \frac{1}{y} \cdot \frac{dy}{dx}$ (using the Chain Rule, where the outer function is $\ln u$ and the inner is $y$). Differentiate terms on the right side using standard rules and the Chain Rule as needed.
- Solve for $\frac{dy}{dx}$. The differentiated equation will contain the term $\frac{1}{y} \frac{dy}{dx}$ on the left side and an expression involving $x$ and possibly $y$ on the right side. Algebraically isolate $\frac{dy}{dx}$ by multiplying both sides of the equation by $y$.
- Substitute the original expression for $y$. Replace $y$ in the resulting formula for $\frac{dy}{dx}$ with its original expression in terms of $x$, i.e., $f(x)$. The final answer for $\frac{dy}{dx}$ should ideally be expressed solely in terms of $x$, although it might sometimes be left in terms of both $x$ and $y$ if appropriate or simpler.
Example Applications
Example 1 (Product/Quotient). Differentiate $y = \frac{\sqrt{x}(x+1)^3}{(2x-1)^4}$.
Answer:
The function involves a product and a quotient of terms raised to powers. Direct application of the Product and Quotient Rules would be complex. Logarithmic differentiation is suitable here.
Step 1: Take natural logarithm of both sides.
Assume $x>0, x+1>0, 2x-1>0$ so $y>0$. (We can use absolute values for full generality, but for most differentiation problems, we focus on intervals where the function is defined and non-zero). Rewrite $\sqrt{x}$ as $x^{1/2}$.
$\ln y = \ln \left( \frac{x^{1/2}(x+1)^3}{(2x-1)^4} \right)$
Step 2: Simplify using logarithm properties.
Use $\ln(A/B) = \ln A - \ln B$ for the main fraction:
$\ln y = \ln(x^{1/2}(x+1)^3) - \ln((2x-1)^4)$
Use $\ln(AB) = \ln A + \ln B$ on the first term and $\ln(A^n) = n \ln A$ on both terms:
$\ln y = \ln(x^{1/2}) + \ln((x+1)^3) - 4\ln(2x-1)$
Use $\ln(A^n) = n \ln A$ on the remaining power terms:
$\ln y = \frac{1}{2}\ln x + 3\ln(x+1) - 4\ln(2x-1)$
This expanded form is much easier to differentiate.
Step 3: Differentiate both sides implicitly with respect to $x$.
Differentiate the left side: $\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}$.
Differentiate the right side term by term. Use the Chain Rule for $\ln(g(x))$, which is $\frac{1}{g(x)} g'(x)$.
$\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}\left(\frac{1}{2}\ln x\right) + \frac{d}{dx}(3\ln(x+1)) - \frac{d}{dx}(4\ln(2x-1))$
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{x} + 3 \cdot \frac{1}{x+1} \cdot \frac{d}{dx}(x+1) - 4 \cdot \frac{1}{2x-1} \cdot \frac{d}{dx}(2x-1)$
Differentiate the inner functions: $\frac{d}{dx}(x+1) = 1$ and $\frac{d}{dx}(2x-1) = 2$.
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2x} + \frac{3}{x+1}(1) - \frac{4}{2x-1}(2)$
$\frac{1}{y} \frac{dy}{dx} = \frac{1}{2x} + \frac{3}{x+1} - \frac{8}{2x-1}$
Step 4: Solve for $\frac{dy}{dx}$.
Multiply both sides of the equation by $y$:
$\frac{dy}{dx} = y \left[ \frac{1}{2x} + \frac{3}{x+1} - \frac{8}{2x-1} \right]$
Step 5: Substitute the original expression for $y$.
Substitute $y = \frac{\sqrt{x}(x+1)^3}{(2x-1)^4}$ back into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\sqrt{x}(x+1)^3}{(2x-1)^4} \left[ \frac{1}{2x} + \frac{3}{x+1} - \frac{8}{2x-1} \right]$
This is the derivative of the given function. While the terms in the bracket could be combined into a single fraction, this form is often acceptable unless further simplification is requested.
Example 2 (Function Power Function). Differentiate $y = x^{\sin x}$ for $x > 0$.
Answer:
This function is of the form $y = [f(x)]^{g(x)}$ where $f(x) = x$ and $g(x) = \sin x$. Since both the base and the exponent are functions of $x$, we must use logarithmic differentiation.
Step 1: Take natural logarithm of both sides.
Since $x > 0$, $y = x^{\sin x}$ will be positive, so we can take the natural logarithm directly.
$\ln y = \ln(x^{\sin x})$
Step 2: Simplify using logarithm properties.
Use the property $\ln(A^n) = n \ln A$ to bring the exponent down as a factor:
$\ln y = (\sin x) (\ln x)$
Now the right side is a product of two functions, $\sin x$ and $\ln x$, which is easy to differentiate using the Product Rule.
Step 3: Differentiate both sides implicitly with respect to $x$.
Differentiate the left side: $\frac{d}{dx}(\ln y) = \frac{1}{y} \frac{dy}{dx}$.
Differentiate the right side using the Product Rule $\frac{d}{dx}(uv) = u v' + v u'$ with $u = \sin x$ and $v = \ln x$.
$u'(x) = \frac{d}{dx}(\sin x) = \cos x$.
$v'(x) = \frac{d}{dx}(\ln x) = \frac{1}{x}$.
Apply the Product Rule to the right side:
$\frac{d}{dx}[(\sin x) (\ln x)] = (\sin x) \cdot \frac{d}{dx}(\ln x) + (\ln x) \cdot \frac{d}{dx}(\sin x)$
$= (\sin x) \cdot \left(\frac{1}{x}\right) + (\ln x) \cdot (\cos x)$
[Substituting individual derivatives]
$= \frac{\sin x}{x} + \cos x \ln x$
Equate the derivatives of the left and right sides:
$\frac{1}{y} \frac{dy}{dx} = \frac{\sin x}{x} + \cos x \ln x$
Step 4: Solve for $\frac{dy}{dx}$.
Multiply both sides of the equation by $y$:
$\frac{dy}{dx} = y \left( \frac{\sin x}{x} + \cos x \ln x \right)$
Step 5: Substitute the original expression for $y$.
Substitute $y = x^{\sin x}$ back into the expression for $\frac{dy}{dx}$:
$\frac{dy}{dx} = x^{\sin x} \left( \frac{\sin x}{x} + \cos x \ln x \right)$
The derivative of $y = x^{\sin x}$ is $x^{\sin x} \left( \frac{\sin x}{x} + \cos x \ln x \right)$.
Functions in Parametric Forms
Not all curves in the Cartesian plane can be easily represented by a single equation $y = f(x)$ or $x = g(y)$, especially if the curve is not a function in $x$ or $y$ (like a circle or a loop), or if we are interested in the motion along the curve over time. In such cases, it is often convenient to describe the coordinates $x$ and $y$ of points on the curve as functions of a third independent variable, called a parameter.
The equations that express $x$ and $y$ in terms of this parameter are called parametric equations.
A curve is represented parametrically by the equations:
$x = f(t)$
$y = g(t)$
Here, $t$ is the parameter. As the parameter $t$ varies over some specified interval (which is the domain of $t$), the point $(x, y) = (f(t), g(t))$ traces out the curve in the Cartesian plane. The interval for $t$ is often specified, but if not, it is usually assumed to be the intersection of the domains of $f$ and $g$.
Parametric equations allow us to describe complex curves, curves that are not functions in $x$ or $y$, and they naturally lend themselves to describing motion, where the parameter $t$ often represents time.
Examples of Functions in Parametric Forms
Many common geometric shapes and paths can be described elegantly using parametric equations:
- Circle centered at the origin with radius $r$:
$x = r \cos t$
$y = r \sin t$
For $0 \le t < 2\pi$. As $t$ goes from 0 to $2\pi$, the point $(x, y)$ starts at $(r, 0)$ and traces out the circle counterclockwise. (We can verify this represents a circle: $x^2 + y^2 = (r \cos t)^2 + (r \sin t)^2 = r^2 \cos^2 t + r^2 \sin^2 t = r^2(\cos^2 t + \sin^2 t) = r^2(1) = r^2$).
- Ellipse centered at the origin with semi-axes $a$ and $b$:
$x = a \cos t$
$y = b \sin t$
For $0 \le t < 2\pi$. This traces an ellipse because $\left(\frac{x}{a}\right)^2 + \left(\frac{y}{b}\right)^2 = \cos^2 t + \sin^2 t = 1$, which is the standard equation of an ellipse.
- Parabola $y^2 = 4ax$:
$x = at^2$
$y = 2at$
For $t \in \mathbb{R}$. If you substitute $t = y/(2a)$ into the equation for $x$, you get $x = a(y/(2a))^2 = a(y^2/(4a^2)) = y^2/(4a)$, so $y^2 = 4ax$. This parametrization covers the entire parabola.
- Line Segment from $(x_1, y_1)$ to $(x_2, y_2)$:
$x = x_1 + t(x_2 - x_1)$
$y = y_1 + t(y_2 - y_1)$
For $0 \le t \le 1$. When $t=0$, $(x, y) = (x_1, y_1)$. When $t=1$, $(x, y) = (x_2, y_2)$. For values of $t$ between 0 and 1, $(x,y)$ lies on the line segment connecting these points.
- Cycloid (Path traced by a point on the circumference of a circle rolling along a straight line):
$x = r(t - \sin t)$
$y = r(1 - \cos t)$
Where $r$ is the radius of the rolling circle and $t$ is the angle through which the circle has rolled. This is a classic example where a direct Cartesian equation is very complex, but parametric equations are relatively simple.
Working with parametric equations requires techniques for finding derivatives $\frac{dy}{dx}$ that are different from those used for explicit or implicit functions.
Derivatives of Functions in Parametric Forms
When a curve is defined by parametric equations, $x = f(t)$ and $y = g(t)$, where $t$ is the parameter, we often need to find the slope of the tangent line to the curve at a specific point. The slope of the tangent line in Cartesian coordinates is given by $\frac{dy}{dx}$. Although we don't have $y$ directly as a function of $x$, we can find $\frac{dy}{dx}$ by using the derivatives of $x$ and $y$ with respect to the parameter $t$ and applying the Chain Rule.
Formula for $\frac{dy}{dx}$ from Parametric Equations
Assume that $x = f(t)$ and $y = g(t)$ are differentiable functions of $t$, and that $f'(t) = \frac{dx}{dt} \neq 0$ over the interval of interest. If $x = f(t)$ can be inverted to express $t$ as a function of $x$, say $t = h(x)$, then $y$ can be viewed as a composite function of $x$: $y = g(t) = g(h(x))$.
We can use the Chain Rule to find the derivative of $y$ with respect to $x$:
$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}$
[Using Chain Rule]
We know $\frac{dy}{dt} = g'(t)$. From the inverse function theorem, $\frac{dt}{dx} = \frac{1}{dx/dt}$ (provided $\frac{dx}{dt} \neq 0$). So, $\frac{dt}{dx} = \frac{1}{f'(t)}$.
Substitute these into the Chain Rule formula:
$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{1}{dx/dt}$
This gives the fundamental formula for the derivative of a parametrically defined function:
$\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$ (provided $\frac{dx}{dt} \neq 0$)
This formula states that the slope of the tangent line $\frac{dy}{dx}$ at a point on the parametric curve is found by dividing the derivative of $y$ with respect to $t$ by the derivative of $x$ with respect to $t$. This allows us to calculate the slope without needing to eliminate the parameter $t$ to find an explicit relationship between $x$ and $y$. The result $\frac{dy}{dx}$ will be expressed in terms of the parameter $t$.
Steps for Finding $\frac{dy}{dx}$ from Parametric Equations
Given parametric equations $x = f(t)$ and $y = g(t)$, follow these steps to find $\frac{dy}{dx}$:
- Identify the given parametric equations for $x$ and $y$ in terms of the parameter $t$.
- Find the derivative of the $x$ equation with respect to $t$. Calculate $\frac{dx}{dt} = f'(t)$.
- Find the derivative of the $y$ equation with respect to $t$. Calculate $\frac{dy}{dt} = g'(t)$.
- Divide the derivative of $y$ by the derivative of $x$ (both with respect to $t$) to find $\frac{dy}{dx}$. Form the ratio $\frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{g'(t)}{f'(t)}$, being careful to note any values of $t$ where $f'(t) = 0$ (where the tangent might be vertical).
To find the slope at a specific point $(x_0, y_0)$ on the curve, you first need to find the value(s) of $t$ that correspond to that point by solving $x_0 = f(t)$ and $y_0 = g(t)$, and then substitute that value of $t$ into the expression for $\frac{dy}{dx}$.
Example 1. Find $\frac{dy}{dx}$ if $x = at^2$ and $y = 2at$, where $a$ is a constant.
Answer:
We are given the parametric equations:
"$x = at^2$"
... (i)
"$y = 2at$"
... (ii)
We need to find $\frac{dy}{dx}$ using the formula $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Step 1: Find the derivative of $x$ with respect to $t$.
Differentiate equation (i) with respect to $t$:
$\frac{dx}{dt} = \frac{d}{dt}(at^2)$
Using the Constant Multiple Rule and Power Rule:
$\frac{dx}{dt} = a \cdot \frac{d}{dt}(t^2) = a(2t^{2-1}) = 2at$
$\frac{dx}{dt} = 2at$
... (iii)
Step 2: Find the derivative of $y$ with respect to $t$.
Differentiate equation (ii) with respect to $t$:
$\frac{dy}{dt} = \frac{d}{dt}(2at)$
Using the Constant Multiple Rule and the fact that $\frac{d}{dt}(t) = 1$:
$\frac{dy}{dt} = 2a \cdot \frac{d}{dt}(t) = 2a(1) = 2a$
$\frac{dy}{dt} = 2a$
... (iv)
Step 3: Calculate $\frac{dy}{dx} = \frac{dy/dt}{dx/dt}$.
Substitute the results from equations (iii) and (iv) into the formula:
$\frac{dy}{dx} = \frac{2a}{2at}$
[Using formula]
Simplify the expression. We can cancel $2a$ from the numerator and denominator, provided $2a \neq 0$ (i.e., $a \neq 0$) and $t \neq 0$ (to avoid division by zero).
$\frac{dy}{dx} = \frac{\cancel{2a}}{\cancel{2a}t} = \frac{1}{t}$
[Simplifying, for $a \neq 0, t \neq 0$]
The derivative $\frac{dy}{dx}$ in terms of the parameter $t$ is $\frac{1}{t}$.
This is the parametric form of a parabola $y^2=4ax$. The derivative $\frac{dy}{dx}$ can also be found explicitly: $y = \pm 2\sqrt{ax}$, so $\frac{dy}{dx} = \pm 2 \cdot \frac{1}{2\sqrt{ax}} \cdot a = \pm \frac{a}{\sqrt{ax}} = \pm \frac{\sqrt{a}}{\sqrt{x}}$. Since $x = at^2$ and $y=2at$, the upper half ($t>0$) corresponds to $y>0$ and $\frac{dy}{dx} = \frac{1}{t} = \frac{2a}{y} = \frac{2a}{2\sqrt{ax}} = \frac{a}{\sqrt{ax}} = \sqrt{\frac{a}{x}}$. The lower half ($t<0$) corresponds to $y<0$ and $\frac{dy}{dx} = \frac{1}{t} = \frac{2a}{y} = \frac{2a}{-2\sqrt{ax}} = -\frac{a}{\sqrt{ax}} = -\sqrt{\frac{a}{x}}$. The formula $\frac{dy}{dx} = \frac{1}{t}$ holds for $t \neq 0$. When $t=0$, $x=0, y=0$, the tangent is vertical and the derivative is undefined.
Example 2. Find $\frac{dy}{dx}$ if $x = a(\theta - \sin \theta)$ and $y = a(1 - \cos \theta)$, where $a$ is a constant.
Answer:
We are given the parametric equations in terms of the parameter $\theta$:
"$x = a(\theta - \sin \theta)$"
... (v)
"$y = a(1 - \cos \theta)$"
... (vi)
We need to find $\frac{dy}{dx}$ using the formula $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.
Step 1: Find the derivative of $x$ with respect to $\theta$.
Differentiate equation (v) with respect to $\theta$:
$\frac{dx}{d\theta} = \frac{d}{d\theta}[a(\theta - \sin \theta)]$
Using the Constant Multiple Rule and Difference Rule:
$= a \frac{d}{d\theta}(\theta - \sin \theta) = a \left(\frac{d}{d\theta}(\theta) - \frac{d}{d\theta}(\sin \theta)\right)$
Using standard derivatives $\frac{d}{d\theta}(\theta)=1$ and $\frac{d}{d\theta}(\sin \theta) = \cos \theta$:
$= a(1 - \cos \theta)$
$\frac{dx}{d\theta} = a(1 - \cos \theta)$
... (vii)
Step 2: Find the derivative of $y$ with respect to $\theta$.
Differentiate equation (vi) with respect to $\theta$:
$\frac{dy}{d\theta} = \frac{d}{d\theta}[a(1 - \cos \theta)]$
Using the Constant Multiple Rule and Difference Rule:
$= a \frac{d}{d\theta}(1 - \cos \theta) = a \left(\frac{d}{d\theta}(1) - \frac{d}{d\theta}(\cos \theta)\right)$
Using standard derivatives $\frac{d}{d\theta}(1)=0$ and $\frac{d}{d\theta}(\cos \theta) = -\sin \theta$:
$= a(0 - (-\sin \theta)) = a(\sin \theta)$
$\frac{dy}{d\theta} = a \sin \theta$
... (viii)
Step 3: Calculate $\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta}$.
Substitute the results from equations (vii) and (viii) into the formula:
$\frac{dy}{dx} = \frac{a \sin \theta}{a(1 - \cos \theta)}$
[Using formula]
Simplify the expression. We can cancel $a$ from the numerator and denominator, provided $a \neq 0$ and $1 - \cos \theta \neq 0$ (i.e., $\cos \theta \neq 1$, which means $\theta \neq 2n\pi$ for integer $n$).
$\frac{dy}{dx} = \frac{\cancel{a} \sin \theta}{\cancel{a}(1 - \cos \theta)} = \frac{\sin \theta}{1 - \cos \theta}$
[Simplifying, for $a \neq 0, \theta \neq 2n\pi$]
This is the derivative in terms of $\theta$. We can simplify this further using half-angle trigonometric identities:
Recall $\sin \theta = 2 \sin(\theta/2) \cos(\theta/2)$ and $1 - \cos \theta = 2 \sin^2(\theta/2)$.
$\frac{dy}{dx} = \frac{2 \sin(\theta/2) \cos(\theta/2)}{2 \sin^2(\theta/2)}$
[Using half-angle identities]
Cancel the common factor $2 \sin(\theta/2)$, provided $\sin(\theta/2) \neq 0$ (i.e., $\theta/2 \neq n\pi$, so $\theta \neq 2n\pi$):
$= \frac{\cancel{2 \sin(\theta/2)} \cos(\theta/2)}{\cancel{2 \sin(\theta/2)} \sin(\theta/2)}$
$= \frac{\cos(\theta/2)}{\sin(\theta/2)} = \cot(\theta/2)$
[Using $\cot u = \cos u / \sin u$]
The derivative $\frac{dy}{dx}$ is $\cot(\theta/2)$ (or equivalently $\frac{\sin \theta}{1 - \cos \theta}$), for values of $\theta$ where $a \neq 0$ and $\theta \neq 2n\pi$. When $\theta = 2n\pi$, $\frac{dx}{d\theta} = 0$, corresponding to points where the cycloid has a vertical tangent (cusps).
Second Derivative in Parametric Form ($\frac{d^2y}{dx^2}$)
To find the second derivative, $\frac{d^2y}{dx^2}$, for a curve defined parametrically, we cannot simply differentiate $\frac{dy}{dx}$ with respect to $t$. The notation $\frac{d^2y}{dx^2}$ means $\frac{d}{dx}\left(\frac{dy}{dx}\right)$. Since $\frac{dy}{dx}$ is typically expressed as a function of the parameter $t$, say $\frac{dy}{dx} = h(t)$, we need to differentiate $h(t)$ with respect to $x$. Again, we use the Chain Rule:
$\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dt}\left(\frac{dy}{dx}\right) \cdot \frac{dt}{dx}$
[Using Chain Rule]
Using the fact that $\frac{dt}{dx} = \frac{1}{dx/dt}$ (assuming $\frac{dx}{dt} \neq 0$), we get the formula for the second derivative:
$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$
Steps for Finding $\frac{d^2y}{dx^2}$ from Parametric Equations:
- Find the first derivative $\frac{dy}{dx}$ as a function of the parameter $t$. Let this function be $h(t)$. (This is Step 1-3 from finding the first derivative).
- Find the derivative of this first derivative with respect to the parameter $t$. Calculate $\frac{d}{dt}\left(\frac{dy}{dx}\right) = \frac{dh}{dt} = h'(t)$.
- Find the derivative of $x$ with respect to $t$, $\frac{dx}{dt}$. (This would have already been calculated when finding the first derivative).
- Divide the result from Step 2 by the result from Step 3: $\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}\left(\frac{dy}{dx}\right)}{\frac{dx}{dt}}$.
The second derivative $\frac{d^2y}{dx^2}$ will also be expressed in terms of the parameter $t$. This gives information about the concavity of the curve.