4.2 Other techniques

4.2.1 Implicit Differentiation

The equation of a circle {x}^{2} + {y}^{2} = {a}^{2} is not in the form y = f(x), (although it can be rearranged to y = ±\sqrt{{a}^{2 } − {x}^{2}}). In this case it is easier to find {dy\over dx} directly without rearranging. Differentiate both sides of the equation {x}^{2} + {y}^{2} = {a}^{2} with respect to x, assuming y to be a function of x. We find 2x + {d{y}^{2}\over dx} = 0. Now use {d({y}^{2})\over dx} = 2y{dy\over dx}. (Proof: Put z = {y}^{2} - need {dz\over dx}, {dz\over dx} = {dz\over dy} {dy\over dx} = 2y{dy\over dx}.) So 2x + 2y{dy\over dx} = 0, or

N.B.: This method usually gives {dy\over dx} in terms of both x and y.

Example 4.9: 

Find {dy\over dx} for {x}^{2} + 4x + 3xy + {y}^{3} = 6.

Solution: 

Differentiating both sides with respect to x we find

2x + 4 + 3y + 3x{dy\over dx} + 3{y}^{2} {dy\over dx} = 0\quad ,

we thus conclude that

\class{boxed}{{dy\over dx} = −{(2x + 4 + 3y)\over (3x + 3{y}^{2})} \quad . }

4.2.2 Logarithmic differentiation

If a function has a large number of factors it may be easier to take the logarithm before differentiating, using the fat that the logarithm of a product is the sum of logarithms.

Example 4.10: 

Find {dy\over dx} for y = {\sqrt{a+x}\sqrt{b−x}\over x−c} .

Solution: 

\mathop{ln}\nolimits y =\mathop{ ln}\nolimits (\sqrt{a + x}) +\mathop{ ln}\nolimits (\sqrt{b − x}) −\mathop{ ln}\nolimits (x − c) = {1\over 2}\mathop{ln}\nolimits (a + x) + {1\over 2}\mathop{ln}\nolimits (b − x) −\mathop{ ln}\nolimits (x − c)\quad .

Differentiate both sides with respect to x:

{d\mathop{ln}\nolimits y\over dx} = {1\over y} {dy\over dx}\quad .

So

{1\over y} {dy\over dx} = {1\over 2} {1\over (a + x)} + {1\over 2} {(−1)\over (b − x)} − {1\over (x − c)}\quad .

and thus

\class{boxed}{{dy\over dx} = {1\over 2}\left ( {1\over (a + x)} − {1\over (b − x)} − {2\over (x − c)}\right ){\sqrt{a + x}\sqrt{b − x}\over x − c} \quad . }

4.2.3 Differentiation of parametric equations

Some equations can be written in parametric form, i.e., x = x(t), y = y(t) with t a parameter. We can then find its differential in terms of the parameter. We shall study this by means of an example only.

Example 4.11: 

Given circle of radius 4,

{x}^{2} + {y}^{2} = 16

(4.1)

use the parametric form to find {dy\over dx} and {{d}^{2}y\over d{x}^{2}} at (2\sqrt{3},2).

Solution: 

The parametric form is

x = 4\mathop{cos}\nolimits θ,\quad y = 4\mathop{sin}\nolimits θ\quad ,

which clearly satisfies (4.1). Now

\class{boxed}{{dy\over dx} = {dy\over df} {df\over dx} = {{dy\over dθ}\over {dx\over dθ}} = {4\mathop{cos}\nolimits θ\over −4\mathop{sin}\nolimits θ} = −\mathop{cot}\nolimits θ. }

Note: result is in terms of θ. Then y∕4 =\mathop{ sin}\nolimits θ = {1\over 2}, θ = {π\over 6} (must be in first quadrant), and \mathop{cos}\nolimits θ = {\sqrt{3}\over 2} therefore {dy\over dx} = −{{\sqrt{3}\over 2} \over {1\over 2} } = −\sqrt{3}. Now do {{d}^{2}y\over d{x}^{2}} .

Note: {{d}^{2}y\over d{x}^{2}} \mathrel{≠}{{d}^{2}y\over d{θ}^{2}} ∕{{d}^{2}x\over d{θ}^{2}}

\begin{eqnarray*} {{d}^{2}y\over d{x}^{2}}& =& {d\over dx} {dy\over dx} %& \\ & =& {d\over dx}(−\mathop{cot}\nolimits θ) = {d\over dθ}(−\mathop{cot}\nolimits θ){dθ\over dx} %& \\ & =& {{\mathop{cosec}\nolimits }^{2}θ\over dx∕dθ} = ({\mathop{cosec}\nolimits }^{2}θ)∕(−4\mathop{sin}\nolimits θ)%& \\ & =& −(1∕4){\mathop{cosec}\nolimits }^{3}θ. %& \\ \end{eqnarray*}

Other examples of parametric curves are

1. Ellipses {x}^{2}∕{a}^{2} + {y}^{2}∕{b}^{2} = 1: put x = a\mathop{cos}\nolimits θ and y = b\mathop{sin}\nolimits θ,
2. Parabola {x}^{2}∕{a}^{2} − {y}^{2}∕{b}^{2} = 1: put x = a\mathop{cosh}\nolimits θ and y = b\mathop{sinh}\nolimits θ.
3. Use of time t, e.g., for x = 2t + 1, y = −g{t}^{2}∕2 + 3t.