Continuity and Differentiability • Topic 3 of 3

Implicit, Logarithmic & Parametric Differentiation

Some relations are not given as $y=f(x)$. Three techniques extend differentiation to these cases.

Implicit differentiation

When $x$ and $y$ are tangled in one equation, differentiate both sides with respect to $x$, treating $y$ as a function of $x$ (so $\tfrac{d}{dx}y^n = ny^{n-1}\tfrac{dy}{dx}$), then solve for $\tfrac{dy}{dx}$.

Logarithmic differentiation

For products/quotients of many factors, or for a variable raised to a variable power such as $y=x^{x}$, take $\ln$ of both sides first. This converts powers into products and products into sums, which are far easier to differentiate. For $y=u(x)^{v(x)}$ this is the only elementary route.

Parametric differentiation

If $x=f(t)$ and $y=g(t)$, then

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{g'(t)}{f'(t)},\qquad f'(t)\ne0.$$

Second-order derivatives

Differentiating $\tfrac{dy}{dx}$ again gives $\tfrac{d^2y}{dx^2}$, the rate of change of the slope — needed for concavity and the second-derivative test in the next chapter.

Solved Examples
1
Worked Example
Find $\dfrac{dy}{dx}$ if $x^2+y^2=25$.
Solution

Differentiate implicitly: $2x+2y\dfrac{dy}{dx}=0 \Rightarrow \dfrac{dy}{dx}=-\dfrac{x}{y}.$

2
Worked Example
Differentiate $y=x^{x}$ ($x>0$).
Solution

Take logs: $\ln y=x\ln x$. Differentiate: $\dfrac{1}{y}\dfrac{dy}{dx}=\ln x+1$. So $\dfrac{dy}{dx}=x^{x}(1+\ln x).$

3
Worked Example
If $x=at^2,\ y=2at$, find $\dfrac{dy}{dx}$.
Solution

$\dfrac{dx}{dt}=2at,\ \dfrac{dy}{dt}=2a$. Hence $\dfrac{dy}{dx}=\dfrac{2a}{2at}=\dfrac{1}{t}.$

4
Worked Example
Find $\dfrac{d^2y}{dx^2}$ for $y=x^3$.
Solution

$\dfrac{dy}{dx}=3x^2$, then $\dfrac{d^2y}{dx^2}=6x.$

Key Points

  • Implicit: differentiate both sides w.r.t. $x$ (use chain rule on $y$), then solve for $\tfrac{dy}{dx}$.
  • Logarithmic differentiation is essential for $y=u(x)^{v(x)}$; take $\ln$ first.
  • Parametric: $\dfrac{dy}{dx}=\dfrac{dy/dt}{dx/dt}$.
  • Second derivative $\tfrac{d^2y}{dx^2}$ is the derivative of the first derivative.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.For $x^2+y^2=25$, $\dfrac{dy}{dx}=$
Explanation: $2x+2yy'=0\Rightarrow y'=-x/y$.
Q2.$\dfrac{d}{dx}x^{x}=$
Explanation: Logarithmic differentiation gives $x^{x}(1+\ln x)$.
Q3.If $x=at^2,y=2at$ then $\dfrac{dy}{dx}=$
Explanation: $(2a)/(2at)=1/t$.
Q4.The second derivative of $y=x^3$ is:
Explanation: $y'=3x^2,\ y''=6x$.
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