← Back to all lessons

Class 12Calculus9:05Published 3 Mar 2026

Second Derivative of a Parametric Function

Two short Class 12 derivative questions: a logarithmic differentiation proof and the second derivative of a parametric function.

This lesson works through two important short-answer derivative questions. The first uses logarithmic differentiation to handle an equation where the variable appears in the exponent, then proves the derivative and evaluates it at a point. The second shows how to find a second derivative when both variables are written in terms of a parameter, using a half-angle simplification to reach a clean result.

What you'll learn

Using logarithmic differentiation when a variable sits in the exponent
Applying the quotient rule and evaluating a derivative at a specific point
Finding the second derivative of a function given in parametric form

Lesson chapters

0:00Two short derivative questions

0:09Setting up logarithmic differentiation

1:52Quotient rule and simplifying the derivative

4:14Evaluating the derivative at x = e

4:49Parametric function: the method

6:06Working through the second derivative

Lesson notes

This lesson covers two important short-answer derivative questions. The first is a logarithmic differentiation proof, and the second finds the second derivative of a function given in parametric form.

Question 1: Logarithmic differentiation

Given $x^{y} = e^{\,x-y}$ , prove that

$\frac{dy}{dx} = \frac{\log x}{\big(\log(ex)\big)^{2}}$

and find its value at $x = e$ .

Because the variable appears in the exponent, we take logarithms of both sides first.

Taking logs and solving for y

$\log\big(x^{y}\big) = \log\big(e^{\,x-y}\big)$

Using $\log a^{m} = m\log a$ , and $\log e = 1$ :

$y\log x = (x-y)\cdot 1$

Bring the $y$ terms together:

$y\log x + y = x \quad\Longrightarrow\quad y(1 + \log x) = x$

$y = \frac{x}{1 + \log x}$

Differentiating with the quotient rule

For $\dfrac{u}{v}$ , the quotient rule gives $\dfrac{v\,u' - u\,v'}{v^{2}}$ . Here $u = x$ and $v = 1 + \log x$ , so $u' = 1$ and $v' = \tfrac{1}{x}$ :

= \frac{1 + \log x - 1}{(1 + \log x)^{2}} = \frac{\log x}{(1 + \log x)^{2}}$$ Since $1 = \log e$, we can write $1 + \log x = \log e + \log x = \log(ex)$, which gives the required form: $$\frac{dy}{dx} = \frac{\log x}{\big(\log(ex)\big)^{2}}$$ #### Value at x = e $$\left.\frac{dy}{dx}\right|_{x=e} = \frac{\log e}{\big(\log(e\cdot e)\big)^{2}} = \frac{\log e}{\big(\log e^{2}\big)^{2}} = \frac{1}{(2\log e)^{2}} = \frac{1}{2^{2}} = \frac{1}{4}$$ ### Question 2: Second derivative of a parametric function Given $x = a(\theta - \sin\theta)$ and $y = a(1 - \cos\theta)$, find $\dfrac{d^{2}y}{dx^{2}}$. When both variables are written in terms of a parameter, we first find $\dfrac{dy}{dx}$, then differentiate again with respect to the parameter and multiply by $\dfrac{d\theta}{dx}$. #### Differentiating x and y with respect to theta $$\frac{dx}{d\theta} = a(1 - \cos\theta) = 2a\sin^{2}\tfrac{\theta}{2}$$ $$\frac{dy}{d\theta} = a\sin\theta = 2a\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}$$ #### First derivative $$\frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = \frac{2a\sin\tfrac{\theta}{2}\cos\tfrac{\theta}{2}}{2a\sin^{2}\tfrac{\theta}{2}} = \frac{\cos\tfrac{\theta}{2}}{\sin\tfrac{\theta}{2}} = \cot\tfrac{\theta}{2}$$ #### Second derivative Differentiate $\cot\tfrac{\theta}{2}$ with respect to $\theta$, then multiply by $\dfrac{d\theta}{dx} = \dfrac{1}{2a\sin^{2}\tfrac{\theta}{2}}$: $$\frac{d^{2}y}{dx^{2}} = \frac{d}{d\theta}\left(\cot\tfrac{\theta}{2}\right)\cdot \frac{d\theta}{dx} = \left(-\tfrac{1}{2}\csc^{2}\tfrac{\theta}{2}\right)\cdot \frac{1}{2a\sin^{2}\tfrac{\theta}{2}}$$ Since $\dfrac{1}{\sin^{2}\tfrac{\theta}{2}} = \csc^{2}\tfrac{\theta}{2}$: $$\frac{d^{2}y}{dx^{2}} = -\frac{1}{4a}\csc^{4}\tfrac{\theta}{2}$$ ### Key takeaways - When a variable appears in an exponent, take logs of both sides first, then differentiate. - The quotient rule is $\dfrac{d}{dx}\left(\dfrac{u}{v}\right) = \dfrac{v\,u' - u\,v'}{v^{2}}$. - For a parametric function, $\dfrac{d^{2}y}{dx^{2}} = \dfrac{d}{d\theta}\!\left(\dfrac{dy}{dx}\right)\cdot \dfrac{d\theta}{dx}$, not simply differentiating twice with respect to $x$.